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IB Maths Analysis & Approaches · Standard Level · Topic 1

Number and algebra

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1AASL1-q00011.1 Standard form Paper 1Easy4 marks

Let $a = 3.6\times10^{5}$ and $b = 1.2\times10^{-2}$. Give each answer in the form $k\times10^{n}$, where $1\le k\lt 10$ and $n\in\mathbb{Z}$.

(a)

Find $ab$.

(2)

(b)

Find $\dfrac{a}{b}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$3.6\times1.2=4.32$ and $10^{5}\times10^{-2}=10^{3}$	(M1)	Award for either product of the decimal parts or correct addition of indices.
$4.32\times10^{3}$	A1
(b)
$3.6\div1.2=3$ and $10^{5}\div10^{-2}=10^{7}$	(M1)
$3\times10^{7}$	A1	Accept $3.0\times10^{7}$. Award A0 for $30\times10^{6}$ or calculator notation such as 3E7.

Answers: (a) $4.32\times10^{3}$ · (b) $3\times10^{7}$

+ Add to test✓ In your test2AASL1-q00021.1 Standard form Paper 1Easy4 marks

(a)

Write $0.000\,407$ in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(1)

(b)

Write $52.3\times10^{4}$ in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(1)

(c)

Write the following numbers in order of size, starting with the smallest.
$3.1\times10^{-2}\qquad 0.0298\qquad 31.2\times10^{-3}\qquad 2.99\times10^{-2}$

(2)

Show mark scheme

Working	Mark	Notes
(a)
$4.07\times10^{-4}$	A1
(b)
$5.23\times10^{5}$	A1
(c)
writing all four in a common form, e.g. $0.031,\ 0.0298,\ 0.0312,\ 0.0299$	(M1)	Or all in the form $a\times10^{-2}$.
$0.0298,\ \ 2.99\times10^{-2},\ \ 3.1\times10^{-2},\ \ 31.2\times10^{-3}$	A1

Answers: (a) $4.07\times10^{-4}$ · (b) $5.23\times10^{5}$ · (c) $0.0298,\ 2.99\times10^{-2},\ 3.1\times10^{-2},\ 31.2\times10^{-3}$

+ Add to test✓ In your test3AASL1-q00031.1 Standard form Paper 1Easy6 marks

Let $a=6.4\times10^{5}$ and $b=3.8\times10^{4}$. Give each answer in the form $k\times10^{n}$, where $1\le k\lt 10$ and $n\in\mathbb{Z}$.

(a)

Find $a+b$.

(2)

(b)

Find $a-10b$.

(2)

(c)

Find $2a^{2}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$640\,000+38\,000$ or $6.4\times10^{5}+0.38\times10^{5}$	(M1)	Award for writing both numbers with the same power of 10 or as ordinary numbers.
$6.78\times10^{5}$	A1
(b)
$10b=3.8\times10^{5}$, so $(6.4-3.8)\times10^{5}$	(M1)
$2.6\times10^{5}$	A1
(c)
$a^{2}=40.96\times10^{10}$	(M1)	Or $2\times6.4^{2}=81.92$.
$8.192\times10^{11}$	A1	Award A0 for $81.92\times10^{10}$ (not in the required form).

Answers: (a) $6.78\times10^{5}$ · (b) $2.6\times10^{5}$ · (c) $8.192\times10^{11}$

+ Add to test✓ In your test4AASL1-q00081.2 Arithmetic sequences and series Paper 1Easy5 marks

An arithmetic sequence has fourth term $u_{4}=17$ and ninth term $u_{9}=42$.

(a)

Find the common difference, $d$.

(2)

(b)

Find the first term, $u_{1}$.

(1)

(c)

Find the sum of the first 20 terms of the sequence.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$u_{1}+3d=17$ and $u_{1}+8d=42$, or $5d=42-17$	(M1)
$d=5$	A1
(b)
$u_{1}=17-3(5)=2$	A1	FT from their $d$.
(c)
$S_{20}=\dfrac{20}{2}\big(2(2)+19(5)\big)$	(M1)	Correct substitution of their values into either sum formula.
$990$	A1

Answers: (a) $d=5$ · (b) $u_{1}=2$ · (c) $S_{20}=990$

+ Add to test✓ In your test5AASL1-q00101.2 Arithmetic sequences and series Paper 1Easy7 marks

Consider the arithmetic sequence $23,\ 17,\ 11,\ 5,\ \ldots$

(a)

Write down the common difference.

(1)

(b)

Find an expression for $u_{n}$ in the form $an+b$, where $a,b\in\mathbb{Z}$.

(2)

(c)

Find the value of $n$ for which $u_{n}=-121$.

(2)

(d)

Determine whether $-100$ is a term of the sequence.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$d=-6$	A1
(b)
$u_{n}=23+(n-1)(-6)$	(M1)
$u_{n}=29-6n$	A1
(c)
$29-6n=-121$	(M1)	FT from (b).
$n=25$	A1
(d)
$29-6n=-100\Rightarrow n=21.5$	M1
$n$ is not an integer, so $-100$ is not a term	R1	Do not award R1 without a valid method.

Answers: (a) $-6$ · (b) $u_{n}=29-6n$ · (c) $n=25$ · (d) not a term

+ Add to test✓ In your test6AASL1-q00141.2 Arithmetic sequences and series Paper 1Easy6 marks

Mira deposits $4000 in an account that pays simple interest at a fixed annual rate. The value of the account at the end of each year forms an arithmetic sequence. After 6 years, the value of the account is $4780.

(a)

Find the interest paid each year.

(2)

(b)

Find the annual rate of interest.

(2)

(c)

Find the number of complete years after which the value of the account first exceeds $6000.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$4000+6d=4780$	(M1)
$130	A1
(b)
$\dfrac{130}{4000}\times100$	(M1)	FT from (a).
$3.25\%$	A1
(c)
$4000+130n\gt 6000\Rightarrow n\gt 15.38\ldots$	(M1)
16 years	A1

Answers: (a) $130 · (b) 3.25% · (c) 16 years

+ Add to test✓ In your test7AASL1-q00151.2 Arithmetic sequences and series Paper 1Easy6 marks

A machine is bought for $48 000. Its value decreases by the same amount at the end of each year. After 3 years the machine is worth $37 200.

(a)

Find the amount by which the value decreases each year.

(2)

(b)

Find the value of the machine after 10 years.

(1)

(c)

Find the number of complete years after which the value of the machine is first less than 10% of the price paid for it.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{48\,000-37\,200}{3}$	(M1)
$3600	A1
(b)
$48\,000-10(3600)=12\,000$, so $12 000	A1	FT from (a).
(c)
10% of the price is $4800	(A1)
$48\,000-3600n\lt 4800\Rightarrow n\gt 12$	(M1)	After 12 years the value is exactly $4800.
13 years	A1	Award A0 for 12.

Answers: (a) $3600 · (b) $12 000 · (c) 13 years

+ Add to test✓ In your test8AASL1-q00231.3 Geometric sequences and series Paper 1Easy7 marks

Consider the geometric sequence $2187,\ 729,\ 243,\ 81,\ \ldots$

(a)

Write down the common ratio.

(1)

(b)

Find $u_{8}$.

(2)

(c)

Given that $2187=3^{7}$, show that $u_{n}=3^{8-n}$.

(2)

(d)

Hence find the value of $n$ for which $u_{n}=\dfrac{1}{81}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$r=\dfrac{1}{3}$	A1
(b)
$2187\left(\dfrac{1}{3}\right)^{7}$	(M1)
$u_{8}=1$	A1
(c)
$u_{n}=3^{7}\left(\dfrac{1}{3}\right)^{n-1}$	M1
$3^{7}\times3^{-(n-1)}=3^{7-n+1}$	A1
$u_{n}=3^{8-n}$	AG
(d)
$3^{8-n}=3^{-4}$	(M1)
$n=12$	A1

Answers: (a) $\tfrac{1}{3}$ · (b) 1 · (d) $n=12$

+ Add to test✓ In your test9AASL1-q00341.4 Financial applications Paper 1Easy5 marks

(a)

An amount of €5000 is invested at an interest rate of 10% per year, compounded annually. Find the value of the investment after 2 years.

(2)

(b)

A boat is bought for €20 000. It depreciates by 25% each year. Find the value of the boat after 2 years.

(2)

(c)

Hence find the total percentage decrease in the value of the boat over the 2 years.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$5000\times1.1^{2}$	(M1)
€6050	A1
(b)
$20\,000\times0.75^{2}$	(M1)	Award M0 for $20\,000\times0.5$.
€11 250	A1
(c)
$\dfrac{20\,000-11\,250}{20\,000}\times100=43.75\%$	A1	FT from (b). Not 50%.

Answers: (a) €6050 · (b) €11 250 · (c) 43.75%

+ Add to test✓ In your test10AASL1-q00451.5 Exponents and introduction to logarithms Paper 1Easy4 marks

Consider the expression $\dfrac{(2x^{3}y^{-2})^{4}}{8x^{5}y^{-3}}$, where $x,y\neq0$.

(a)

Simplify the expression, giving your answer in the form $ax^{p}y^{q}$, where $a,p,q\in\mathbb{Z}$.

(3)

(b)

Hence find the value of the expression when $x=2$ and $y=2$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$(2x^{3}y^{-2})^{4}=16x^{12}y^{-8}$	(M1)
$2x^{7}$	A1
$y^{-5}$, giving $2x^{7}y^{-5}$	A1
(b)
$2\times2^{7}\times2^{-5}=8$	A1	FT from their simplified expression.

Answers: (a) $2x^{7}y^{-5}$ · (b) $8$

+ Add to test✓ In your test11AASL1-q00461.5 Exponents and introduction to logarithms Paper 1Easy4 marks

(a)

Write down the value of $\log_{10}0.001$.

(1)

(b)

Write down the value of $\ln\dfrac{1}{\mathrm{e}^{2}}$.

(1)

(c)

Solve $10^{2x-1}=1000$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$-3$	A1
(b)
$-2$	A1
(c)
$2x-1=3$, or $2x-1=\log_{10}1000$	(M1)
$x=2$	A1

Answers: (a) $-3$ · (b) $-2$ · (c) $x=2$

+ Add to test✓ In your test12AASL1-q00471.5 Exponents and introduction to logarithms Paper 1Easy5 marks

(a)

Write the equation $10^{-2}=0.01$ in logarithmic form.

(1)

(b)

Given that $\ln y=3$, write down the value of $y$ as a power of $\mathrm{e}$.

(1)

(c)

Given that $\log_{a}81=4$, where $a\gt 0$, $a\neq1$, find the value of $a$.

(2)

(d)

Explain why there is no real value of $x$ for which $10^{x}=-4$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$\log_{10}0.01=-2$	A1	Accept $\lg 0.01=-2$ or $\log 0.01=-2$.
(b)
$y=\mathrm{e}^{3}$	A1
(c)
$a^{4}=81$	(M1)
$a=3$	A1	Award A0 if $a=-3$ is also given.
(d)
$10^{x}\gt 0$ for all real $x$ (equivalently, $\log_{10}b$ is defined only for $b\gt 0$)	R1	Accept any correct reason referring to the positivity of the power or to the domain of the logarithm.

Answers: (a) $\log_{10}0.01=-2$ · (b) $y=\mathrm{e}^{3}$ · (c) $a=3$

+ Add to test✓ In your test13AASL1-q00481.5 Exponents and introduction to logarithms Paper 1Easy4 marks

(a)

Write $\dfrac{8^{-2}\times4^{3}}{2^{-5}}$ in the form $2^{k}$, where $k\in\mathbb{Z}$.

(2)

(b)

Find the value of $n$ for which $5^{n}\times25^{-1}=\dfrac{1}{125}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$8^{-2}=2^{-6}$ and $4^{3}=2^{6}$	(M1)	Award for both conversions to base 2.
$2^{-6+6-(-5)}=2^{5}$	A1
(b)
$5^{n-2}=5^{-3}$	(M1)
$n=-1$	A1

Answers: (a) $2^{5}$ · (b) $n=-1$

+ Add to test✓ In your test14AASL1-q00551.6 Deductive proof Paper 1Easy5 marks

(a)

Show that $8^{2}-7^{2}=8+7$.

(1)

(b)

Show that $(n+1)^{2}-n^{2}\equiv 2n+1$.

(2)

(c)

Hence write 41 as the difference of the squares of two consecutive integers.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$64-49=15$ and $8+7=15$	A1	Both sides must be evaluated.
(b)
$n^{2}+2n+1-n^{2}$	M1	Must start from the LHS.
$2n+1$	A1	Final line must show the simplification, not just the answer.
$(n+1)^{2}-n^{2}\equiv2n+1$	AG
(c)
$2n+1=41$, so $n=20$	(M1)
$41=21^{2}-20^{2}$	A1

Answers: (c) $41=21^{2}-20^{2}$

+ Add to test✓ In your test15AASL1-q00571.6 Deductive proof Paper 1Easy6 marks

A student writes $(x+2)^{2}\equiv x^{2}+4$.

(a)

By substituting $x=1$, show that the student’s statement is not an identity.

(1)

(b)

Find all values of $x$ for which $(x+2)^{2}=x^{2}+4$.

(2)

(c)

Show that $(x+2)^{2}-(x-2)^{2}\equiv8x$.

(2)

(d)

Check your result in part (c) by substituting $x=5$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
LHS $=9$, RHS $=5$, and $9\neq5$	A1
(b)
$x^{2}+4x+4=x^{2}+4$, so $4x=0$	(M1)
$x=0$ only	A1
(c)
$\left(x^{2}+4x+4\right)-\left(x^{2}-4x+4\right)$	M1	Brackets (or correct signs) required.
$=x^{2}+4x+4-x^{2}+4x-4$	A1
$\equiv8x$	AG
(d)
LHS $=49-9=40$; RHS $=8\times5=40$	A1

Answers: (b) $x=0$

+ Add to test✓ In your test16AASL1-q00601.7 Rational exponents and laws of logarithms Paper 1Easy5 marks

(a)

Find the value of $32^{\frac{3}{5}}$.

(1)

(b)

Find the value of $\left(\dfrac{8}{27}\right)^{-\frac{2}{3}}$.

(2)

(c)

Solve $x^{\frac{3}{2}}=64$, for $x\gt 0$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$8$	A1
(b)
$\left(\dfrac{27}{8}\right)^{\frac{2}{3}}$ or $\left(\dfrac{2}{3}\right)^{-2}$	(M1)
$\dfrac{9}{4}$	A1
(c)
$x=64^{\frac{2}{3}}$	(M1)
$x=16$	A1

Answers: (a) 8 · (b) $\tfrac{9}{4}$ · (c) $x=16$

+ Add to test✓ In your test17AASL1-q00641.7 Rational exponents and laws of logarithms Paper 1Easy6 marks

(a)

Find the value of $2\log_{6}3+\log_{6}4$.

(3)

(b)

Write $3\ln2-\ln4+\ln5$ in the form $\ln k$, where $k\in\mathbb{Z}$.

(2)

(c)

Write down the value of $\log_{5}1+\log_{5}5$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$2\log_{6}3=\log_{6}9$	M1
$\log_{6}9+\log_{6}4=\log_{6}36$	M1
$2$	A1
(b)
$\ln8-\ln4+\ln5=\ln\dfrac{8\times5}{4}$	(M1)	Award for at least two correct uses of the laws.
$\ln10$	A1
(c)
$0+1=1$	A1

Answers: (a) 2 · (b) $\ln10$ · (c) 1

+ Add to test✓ In your test18AASL1-q00661.7 Rational exponents and laws of logarithms Paper 1Easy8 marks

The temperature, $T$ °C, of a cup of tea $t$ minutes after it is poured is modelled by $T=20+70\mathrm{e}^{-kt}$, $t\ge0$, where $k$ is a positive constant. After 10 minutes the temperature of the tea is 55 °C.

(a)

Write down the temperature of the tea when it is poured.

(1)

(b)

Show that $k=\dfrac{\ln2}{10}$.

(3)

(c)

Find the time taken for the temperature of the tea to fall to 37.5 °C.

(3)

(d)

Explain why, according to the model, the temperature of the tea never falls to 20 °C.

(1)

Show mark scheme

Working	Mark	Notes
(a)
90 °C	A1
(b)
$20+70\mathrm{e}^{-10k}=55$	M1
$\mathrm{e}^{-10k}=\tfrac{1}{2}$	A1
$-10k=\ln\tfrac{1}{2}=-\ln2$	A1
$k=\dfrac{\ln2}{10}$	AG
(c)
$70\mathrm{e}^{-kt}=17.5$, i.e. $\mathrm{e}^{-kt}=\tfrac{1}{4}$	(M1)
$-\dfrac{\ln2}{10}t=\ln\tfrac{1}{4}=-2\ln2$	(M1)	Or $\mathrm{e}^{-10k}=\tfrac{1}{2}$, so $\tfrac{1}{4}=\left(\mathrm{e}^{-10k}\right)^{2}=\mathrm{e}^{-20k}$.
$t=20$ minutes	A1
(d)
$\mathrm{e}^{-kt}\gt 0$ for all $t$, so $T\gt 20$	R1	Accept $70\mathrm{e}^{-kt}=0$ has no solution.

Answers: (a) 90 °C · (c) 20 minutes

+ Add to test✓ In your test19AASL1-q00741.8 Infinite geometric series Paper 1Easy5 marks

Consider the infinite geometric series $24-12+6-3+\cdots$.

(a)

Write down the common ratio, $r$.

(1)

(b)

Find the sum to infinity of the series.

(2)

(c)

A different infinite geometric series has sum to infinity 45 and common ratio $\tfrac{1}{3}$. Find its first term.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$r=-\tfrac{1}{2}$	A1
(b)
$\dfrac{24}{1-\left(-\frac{1}{2}\right)}$	(M1)
$16$	A1
(c)
$\dfrac{u_{1}}{1-\frac{1}{3}}=45$	(M1)
$u_{1}=30$	A1

Answers: (a) $-\tfrac{1}{2}$ · (b) 16 · (c) 30

+ Add to test✓ In your test20AASL1-q00831.9 The binomial theorem Paper 1Easy5 marks

(a)

Write down the row of Pascal’s triangle that gives the coefficients in the expansion of $(a+b)^{5}$.

(1)

(b)

Use the formula for $\dbinom{n}{r}$ to show that $\dbinom{7}{3}=35$.

(2)

(c)

Write down the value of $\dbinom{7}{4}$.

(1)

(d)

Write down the term in $a^{2}b^{5}$ in the expansion of $(a+b)^{7}$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
1, 5, 10, 10, 5, 1	A1
(b)
$\dfrac{7!}{3!\,4!}$	M1
$=\dfrac{7\times6\times5}{3\times2\times1}=\dfrac{210}{6}$	A1
$=35$	AG
(c)
$35$	A1
(d)
$21a^{2}b^{5}$	A1	Accept $\dbinom{7}{5}a^{2}b^{5}$ only if evaluated.

Answers: (a) 1, 5, 10, 10, 5, 1 · (c) 35 · (d) $21a^{2}b^{5}$

+ Add to test✓ In your test21AASL1-q00041.1 Standard form Paper 1Medium7 marks

Let $p=1.6\times10^{9}$ and $q=4\times10^{-3}$.

(a)

Find $\sqrt{p}$. Give your answer in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(2)

(b)

Find $pq^{2}$. Give your answer in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(2)

(c)

Find the integer $n$ such that $10^{n}\le\dfrac{p}{q^{3}}\lt 10^{n+1}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$p=16\times10^{8}$, so $\sqrt{p}=\sqrt{16}\times\sqrt{10^{8}}$	(M1)
$4\times10^{4}$	A1
(b)
$q^{2}=16\times10^{-6}=1.6\times10^{-5}$	(M1)
$2.56\times10^{4}$	A1
(c)
$q^{3}=64\times10^{-9}=6.4\times10^{-8}$	(M1)
$\dfrac{1.6\times10^{9}}{6.4\times10^{-8}}=0.25\times10^{17}=2.5\times10^{16}$	(A1)
$n=16$	A1

Answers: (a) $4\times10^{4}$ · (b) $2.56\times10^{4}$ · (c) $n=16$

+ Add to test✓ In your test22AASL1-q00091.2 Arithmetic sequences and series Paper 1Medium6 marks

Consider the series $\displaystyle\sum_{k=1}^{n}(4k-1)$, where $n\in\mathbb{Z}^{+}$.

(a)

Write down the first three terms of the series.

(1)

(b)

Show that $\displaystyle\sum_{k=1}^{n}(4k-1)=2n^{2}+n$.

(2)

(c)

Given that $\displaystyle\sum_{k=1}^{n}(4k-1)=210$, find the value of $n$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3+7+11$	A1	Accept 3, 7, 11.
(b)
recognizing an arithmetic series with $u_{1}=3,\ d=4$ and substituting: $\dfrac{n}{2}\big(2(3)+4(n-1)\big)$	M1
$\dfrac{n}{2}(4n+2)$ or $n(2n+1)$	A1
$=2n^{2}+n$	AG
(c)
$2n^{2}+n-210=0$	M1
$(2n+21)(n-10)=0$	A1	Or correct substitution into the quadratic formula.
$n=10$	A1	Award the final A1 only if $n=-\tfrac{21}{2}$ is rejected or not given.

Answers: (a) $3+7+11$ · (c) $n=10$

+ Add to test✓ In your test23AASL1-q00111.2 Arithmetic sequences and series Paper 1Medium8 marks

The second, third and fourth terms of an arithmetic sequence are $k+5$, $k^{2}$ and $5k+3$, in that order, where $k\in\mathbb{R}$.

(a)

Show that $k^{2}-3k-4=0$.

(2)

(b)

The common difference of the sequence is positive. Find the value of $k$.

(3)

(c)

Hence find the sum of the first 20 terms of the sequence.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$k^{2}-(k+5)=(5k+3)-k^{2}$, or $2k^{2}=(k+5)+(5k+3)$	M1
$2k^{2}=6k+8$	A1
$k^{2}-3k-4=0$	AG
(b)
$(k-4)(k+1)=0$, so $k=4$ or $k=-1$	A1
$k=-1$ gives terms $4,\ 1,\ -2$ ($d=-3\lt 0$); $k=4$ gives $9,\ 16,\ 23$ ($d=7\gt 0$)	R1	Must consider the sign of $d$ for each value.
$k=4$	A1
(c)
$d=7$ and $u_{1}=9-7=2$	A1
$S_{20}=\dfrac{20}{2}\big(2(2)+19(7)\big)$	(M1)	FT from their $u_{1}$ and $d$.
$1370$	A1

Answers: (b) $k=4$ · (c) $S_{20}=1370$

+ Add to test✓ In your test24AASL1-q00131.2 Arithmetic sequences and series Paper 1Medium6 marks

Consider the arithmetic series $\displaystyle\sum_{r=5}^{40}(3r+2)$.

(a)

Write down the number of terms in the series.

(1)

(b)

Find the sum of the series.

(3)

(c)

Write the series $8+13+18+\cdots+203$ in sigma notation, in the form $\displaystyle\sum_{r=1}^{n}(ar+b)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$36$	A1
(b)
first term $17$, last term $122$	A1	Or $u_{1}=17,\ d=3$.
$\dfrac{36}{2}(17+122)$	(M1)	FT from their number of terms.
$2502$	A1
(c)
$5r+3$	A1
$\displaystyle\sum_{r=1}^{40}(5r+3)$	A1	Upper limit 40 required.

Answers: (a) 36 · (b) 2502 · (c) $\sum_{r=1}^{40}(5r+3)$

+ Add to test✓ In your test25AASL1-q00221.3 Geometric sequences and series Paper 1Medium6 marks

In a geometric sequence, the second term is $12$ and the fifth term is $-324$.

(a)

Find the common ratio, $r$.

(2)

(b)

Find the first term, $u_{1}$.

(1)

(c)

Find the sum of the first six terms of the sequence.

(2)

(d)

Explain why the sum to infinity of this sequence does not exist.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$u_{1}r=12$ and $u_{1}r^{4}=-324$, so $r^{3}=-27$	(M1)
$r=-3$	A1
(b)
$u_{1}=-4$	A1	FT from their $r$.
(c)
$S_{6}=\dfrac{-4\big((-3)^{6}-1\big)}{-3-1}$	(M1)
$728$	A1
(d)
$\|r\|=3\gt 1$, so the series diverges	R1	Must refer to the value of $\|r\|$ or $r$.

Answers: (a) $r=-3$ · (b) $u_{1}=-4$ · (c) $728$

+ Add to test✓ In your test26AASL1-q00241.3 Geometric sequences and series Paper 1Medium7 marks

The first three terms of a geometric sequence are $k-4$, $k-2$ and $2k-7$, where $k\in\mathbb{R}$.

(a)

Show that $k^{2}-11k+24=0$.

(2)

(b)

All the terms of the sequence are positive. Find the value of $k$ and the common ratio.

(3)

(c)

Find the fifth term of the sequence.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{k-2}{k-4}=\dfrac{2k-7}{k-2}$, or $(k-2)^{2}=(k-4)(2k-7)$	M1
$k^{2}-4k+4=2k^{2}-15k+28$	A1
$k^{2}-11k+24=0$	AG
(b)
$(k-3)(k-8)=0$, so $k=3$ or $k=8$	A1
$k=3$ gives $-1,\ 1,\ -1$, which are not all positive	R1
$k=8$, terms $4,\ 6,\ 9$, $r=\dfrac{3}{2}$	A1	Both $k$ and $r$ required.
(c)
$4\left(\dfrac{3}{2}\right)^{4}$	(M1)	FT from (b).
$\dfrac{81}{4}$	A1	Accept 20.25.

Answers: (b) $k=8,\ r=\tfrac{3}{2}$ · (c) $\tfrac{81}{4}$

+ Add to test✓ In your test27AASL1-q00251.3 Geometric sequences and series Paper 1Medium5 marks

Consider the series $\displaystyle\sum_{k=1}^{8}3\left(2^{k}\right)$.

(a)

Write down the first three terms of the series.

(1)

(b)

Find the sum of the series.

(2)

(c)

Hence find the value of $\displaystyle\sum_{k=4}^{8}3\left(2^{k}\right)$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$6+12+24$	A1
(b)
$\dfrac{6\left(2^{8}-1\right)}{2-1}$	(M1)	Recognizes $u_{1}=6,\ r=2,\ n=8$.
$1530$	A1
(c)
$1530-(6+12+24)$	(M1)	Must use (b).
$1488$	A1

Answers: (a) $6+12+24$ · (b) 1530 · (c) 1488

+ Add to test✓ In your test28AASL1-q00491.5 Exponents and introduction to logarithms Paper 1Medium7 marks

(a)

Find the two consecutive integers between which $\log_{10}4500$ lies. Justify your answer.

(2)

(b)

Write down the value of $\mathrm{e}^{\ln 7}$.

(1)

(c)

Solve $\mathrm{e}^{2x}=5$, giving your answer in the form $a\ln b$, where $a,b\in\mathbb{Q}$.

(2)

(d)

Given that $2.7\lt\mathrm{e}\lt 2.8$, show that the solution to part (c) lies between $\tfrac{1}{2}$ and $1$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$10^{3}=1000\lt 4500\lt 10\,000=10^{4}$, so $3\lt\log_{10}4500\lt 4$	R1
3 and 4	A1
(b)
$7$	A1
(c)
$2x=\ln5$	(M1)
$x=\tfrac{1}{2}\ln5$	A1
(d)
$\mathrm{e}\lt 2.8\lt 5$ and $\mathrm{e}^{2}\gt 2.7^{2}=7.29\gt 5$, so $\mathrm{e}^{1}\lt 5\lt\mathrm{e}^{2}$	M1	Must see a numerical comparison of 5 with both e and e².
hence $1\lt\ln5\lt 2$, so $\tfrac{1}{2}\lt\tfrac{1}{2}\ln5\lt 1$	R1

Answers: (a) 3 and 4 · (b) 7 · (c) $x=\tfrac{1}{2}\ln5$

+ Add to test✓ In your test29AASL1-q00531.6 Deductive proof Paper 1Medium4 marks

Let $n\in\mathbb{Z}^{+}$.

(a)

Show that $\dfrac{1}{n}-\dfrac{1}{n+2}\equiv\dfrac{2}{n(n+2)}$.

(2)

(b)

Hence write $\dfrac{2}{63}$ in the form $\dfrac{1}{n}-\dfrac{1}{n+2}$, where $n\in\mathbb{Z}^{+}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{(n+2)-n}{n(n+2)}$	M1	Must start from the LHS and use a common denominator.
numerator simplified to $2$	A1
$\equiv\dfrac{2}{n(n+2)}$	AG
(b)
$n(n+2)=63$, so $n=7$	(M1)
$\dfrac{1}{7}-\dfrac{1}{9}$	A1

Answers: (b) $\dfrac{2}{63}=\dfrac{1}{7}-\dfrac{1}{9}$

+ Add to test✓ In your test30AASL1-q00541.6 Deductive proof Paper 1Medium6 marks

Let $n\in\mathbb{Z}$.

(a)

Show that $(n-1)^{2}+n^{2}+(n+1)^{2}\equiv3n^{2}+2$.

(2)

(b)

Hence explain why the sum of the squares of any three consecutive integers is never a multiple of 3.

(2)

(c)

Find all sets of three consecutive integers whose squares have a sum of 110.

(2)

Show mark scheme

Working	Mark	Notes
(a)
expanding at least two of the brackets correctly	M1
$n^{2}-2n+1+n^{2}+n^{2}+2n+1$	A1
$\equiv3n^{2}+2$	AG
(b)
$3n^{2}$ is a multiple of 3 for every integer $n$	R1
so the sum is always 2 more than a multiple of 3, hence it is never a multiple of 3	R1
(c)
$3n^{2}+2=110$, so $n^{2}=36$	(M1)
5, 6, 7 and $-7,\,-6,\,-5$	A1	Both sets required for A1.

Answers: (c) 5, 6, 7 and −7, −6, −5

+ Add to test✓ In your test31AASL1-q00561.6 Deductive proof Paper 1Medium6 marks

(a)

Show that $2x^{2}-12x+23\equiv2(x-3)^{2}+5$.

(2)

(b)

Hence explain why $2x^{2}-12x+23\gt 0$ for all $x\in\mathbb{R}$.

(2)

(c)

Hence find the greatest value of $\dfrac{10}{2x^{2}-12x+23}$, and the value of $x$ at which it occurs.

(2)

Show mark scheme

Working	Mark	Notes
(a)
RHS $=2\left(x^{2}-6x+9\right)+5$	M1	Working from either side is acceptable; must show the expansion or the completion of the square.
$=2x^{2}-12x+18+5$	A1
$=2x^{2}-12x+23$	AG
(b)
$(x-3)^{2}\ge0$ for all real $x$	R1
so $2(x-3)^{2}+5\ge5\gt 0$	R1
(c)
denominator has least value 5	(M1)
greatest value 2, when $x=3$	A1	Both required.

Answers: (c) greatest value 2, at $x=3$

+ Add to test✓ In your test32AASL1-q00581.6 Deductive proof Paper 1Medium6 marks

Let $n\in\mathbb{Z}$.

(a)

Show that $(2n+1)^{2}\equiv4n(n+1)+1$.

(2)

(b)

Explain why $n(n+1)$ is even.

(1)

(c)

Hence prove that the square of every odd integer is one more than a multiple of 8.

(2)

(d)

Write down the remainder when $2025^{2}$ is divided by 8.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$4n^{2}+4n+1$	M1
$=4\left(n^{2}+n\right)+1$	A1
$\equiv4n(n+1)+1$	AG
(b)
$n$ and $n+1$ are consecutive integers, so one of them is even	R1
(c)
$n(n+1)=2k$ for some $k\in\mathbb{Z}$, so $4n(n+1)=8k$	R1
every odd integer can be written as $2n+1$, so its square is $8k+1$, one more than a multiple of 8	R1
(d)
$1$	A1

Answers: (d) 1

+ Add to test✓ In your test33AASL1-q00611.7 Rational exponents and laws of logarithms Paper 1Medium6 marks

Let $\log_{3}a=p$ and $\log_{3}b=q$, where $a,b\gt 0$. Find an expression, in terms of $p$ and $q$, for

(a)

$\log_{3}(a^{2}b)$;

(2)

(b)

$\log_{3}\left(\dfrac{9a}{\sqrt{b}}\right)$;

(2)

(c)

$\log_{9}b$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\log_{3}a^{2}+\log_{3}b$	(M1)
$2p+q$	A1
(b)
$\log_{3}9+\log_{3}a-\tfrac{1}{2}\log_{3}b$	(M1)	Award for two correct uses of the laws of logarithms.
$2+p-\dfrac{q}{2}$	A1
(c)
$\dfrac{\log_{3}b}{\log_{3}9}$	(M1)
$\dfrac{q}{2}$	A1

Answers: (a) $2p+q$ · (b) $2+p-\tfrac{q}{2}$ · (c) $\tfrac{q}{2}$

+ Add to test✓ In your test34AASL1-q00621.7 Rational exponents and laws of logarithms Paper 1Medium7 marks

(a)

Solve $4^{x+1}=8^{2x-3}$.

(3)

(b)

By using the substitution $y=5^{x}$, solve $5^{2x}-6\left(5^{x}\right)+5=0$.

(4)

Show mark scheme

Working	Mark	Notes
(a)
writing both sides as powers of 2: $2^{2x+2}=2^{6x-9}$	M1
equating the exponents: $2x+2=6x-9$	M1
$x=\dfrac{11}{4}$	A1
(b)
$y^{2}-6y+5=0$	M1
$y=1$ or $y=5$	A1
$5^{x}=1$ or $5^{x}=5$	(M1)
$x=0$ or $x=1$	A1	Both values required.

Answers: (a) $x=\tfrac{11}{4}$ · (b) $x=0$ or $x=1$

+ Add to test✓ In your test35AASL1-q00651.7 Rational exponents and laws of logarithms Paper 1Medium7 marks

(a)

Solve $5^{x}=3^{x+1}$, giving your answer in the form $\dfrac{\ln a}{\ln b}$, where $a,b\in\mathbb{Q}^{+}$.

(4)

(b)

Solve $2^{x}\times3^{x+1}=12^{x}$, giving your answer in the form $\log_{2}c$, where $c\in\mathbb{Z}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
taking logarithms of both sides: $x\ln5=(x+1)\ln3$	M1	Any base.
collecting the terms in $x$: $x(\ln5-\ln3)=\ln3$	M1
$x=\dfrac{\ln3}{\ln5-\ln3}$	A1
$x=\dfrac{\ln3}{\ln\frac{5}{3}}$	A1	Accept $a=3,\ b=\tfrac{5}{3}$.
(b)
$2^{x}\times3^{x}\times3=6^{x}\times3$, so $3\times6^{x}=12^{x}$	M1	Or taking logarithms: $x\ln2+(x+1)\ln3=x\ln12$.
$\left(\dfrac{12}{6}\right)^{x}=3$, i.e. $2^{x}=3$	A1	Or $x(\ln12-\ln2-\ln3)=\ln3$, i.e. $x\ln2=\ln3$.
$x=\log_{2}3$	A1

Answers: (a) $x=\dfrac{\ln3}{\ln\frac{5}{3}}$ · (b) $x=\log_{2}3$

+ Add to test✓ In your test36AASL1-q00671.7 Rational exponents and laws of logarithms Paper 1Medium7 marks

(a)

Write $\dfrac{x\sqrt{x}}{\sqrt[3]{x}}$ in the form $x^{p}$, where $x\gt 0$ and $p\in\mathbb{Q}$.

(2)

(b)

Solve $\sqrt[3]{x^{2}}=4\sqrt{x}$, for $x\gt 0$.

(3)

(c)

Find the value of $k$ for which $\dfrac{\sqrt{8}\times4^{\frac{1}{3}}}{\sqrt[6]{2}}=2^{k}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$x^{\frac{3}{2}}\div x^{\frac{1}{3}}$	(M1)
$x^{\frac{7}{6}}$	A1
(b)
$x^{\frac{2}{3}}=4x^{\frac{1}{2}}$	(M1)
$x^{\frac{2}{3}-\frac{1}{2}}=x^{\frac{1}{6}}=4$	(M1)	Division by x^(1/2) is valid since x > 0.
$x=4^{6}=4096$	A1	Accept $2^{12}$.
(c)
$2^{\frac{3}{2}}\times2^{\frac{2}{3}}\div2^{\frac{1}{6}}$	(M1)
$k=2$	A1

Answers: (a) $x^{\frac{7}{6}}$ · (b) $x=4096$ · (c) $k=2$

+ Add to test✓ In your test37AASL1-q00691.7 Rational exponents and laws of logarithms Paper 1Medium5 marks

Consider the equation $\log_{2}x+\log_{2}(x-6)=4$.

(a)

State the set of values of $x$ for which both logarithms in the equation are defined.

(1)

(b)

Solve the equation.

(4)

Show mark scheme

Working	Mark	Notes
(a)
$x\gt 6$	A1
(b)
$\log_{2}\big(x(x-6)\big)=4$	M1
$x(x-6)=2^{4}$	M1
$x^{2}-6x-16=0$, so $x=8$ or $x=-2$	A1
$x=8$ only	A1	Award the final A1 only if $x=-2$ is rejected (it is not in $x\gt 6$) or not given.

Answers: (a) $x\gt 6$ · (b) $x=8$

+ Add to test✓ In your test38AASL1-q00751.8 Infinite geometric series Paper 1Medium7 marks

The recurring decimal $0.\dot{3}\dot{6}=0.363636\ldots$ can be written as an infinite geometric series.

(a)

Write down the first three terms of this series and its common ratio.

(2)

(b)

Hence write $0.\dot{3}\dot{6}$ as a fraction in its lowest terms.

(2)

(c)

Write $0.1\dot{2}\dot{7}=0.1272727\ldots$ as a fraction in its lowest terms.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$0.36+0.0036+0.000036$	A1
$r=0.01$	A1
(b)
$\dfrac{0.36}{1-0.01}=\dfrac{36}{99}$	(M1)
$\dfrac{4}{11}$	A1
(c)
$0.1+(0.027+0.00027+\cdots)$	M1
$0.1+\dfrac{0.027}{0.99}=\dfrac{1}{10}+\dfrac{27}{990}$	M1	Or $\tfrac{1}{10}\left(1+\tfrac{27}{99}\right)$.
$\dfrac{7}{55}$	A1

Answers: (a) $0.36+0.0036+0.000036$, $r=0.01$ · (b) $\tfrac{4}{11}$ · (c) $\tfrac{7}{55}$

+ Add to test✓ In your test39AASL1-q00801.9 The binomial theorem Paper 1Medium6 marks

(a)

Use Pascal’s triangle to expand $(2-x)^{4}$.

(3)

(b)

Hence find the coefficient of $x^{3}$ in the expansion of $(1+3x)(2-x)^{4}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
binomial coefficients 1, 4, 6, 4, 1	(M1)
at least three terms correct, e.g. $16,\ -32x,\ 24x^{2}$	(A1)
$16-32x+24x^{2}-8x^{3}+x^{4}$	A1
(b)
identifying the two products that give $x^{3}$: $1\times(-8x^{3})$ and $3x\times24x^{2}$	(M1)
$-8+72$	A1	FT from their expansion.
$64$	A1

Answers: (a) $16-32x+24x^{2}-8x^{3}+x^{4}$ · (b) 64

+ Add to test✓ In your test40AASL1-q00841.9 The binomial theorem Paper 1Medium7 marks

(a)

Find the first three terms, in ascending powers of $x$, of the expansion of $\left(1+\dfrac{x}{4}\right)^{8}$.

(3)

(b)

By substituting a suitable value of $x$ into your answer to part (a), find an approximate value of $1.025^{8}$.

(3)

(c)

State, with a reason, whether your answer to part (b) is an overestimate or an underestimate.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$1+\dbinom{8}{1}\left(\dfrac{x}{4}\right)+\dbinom{8}{2}\left(\dfrac{x}{4}\right)^{2}$	(M1)
$1+2x$	A1
$+\dfrac{7}{4}x^{2}$	A1	$28\times\tfrac{1}{16}=\tfrac{7}{4}$
(b)
$\dfrac{x}{4}=0.025$, so $x=0.1$	(M1)
$1+2(0.1)+\dfrac{7}{4}(0.01)$	(M1)
$1.2175$	A1
(c)
underestimate, since all the omitted terms are positive	R1	Must give the reason.

Answers: (a) $1+2x+\tfrac{7}{4}x^{2}$ · (b) 1.2175 · (c) underestimate

+ Add to test✓ In your test41AASL1-q00861.9 The binomial theorem Paper 1Medium7 marks

Let $n\in\mathbb{Z}$, $n\ge2$.

(a)

Show that $\dbinom{n}{2}=\dfrac{n(n-1)}{2}$.

(2)

(b)

Solve $\dbinom{n}{2}=45$.

(3)

(c)

Show that $\dbinom{n}{2}+\dbinom{n+1}{2}\equiv n^{2}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{n!}{2!\,(n-2)!}$	M1
$\dfrac{n(n-1)(n-2)!}{2\,(n-2)!}$	A1
$=\dfrac{n(n-1)}{2}$	AG
(b)
$n(n-1)=90$, i.e. $n^{2}-n-90=0$	M1
$(n-10)(n+9)=0$	A1
$n=10$	A1	Award the final A1 only if $n=-9$ is rejected or not given.
(c)
$\dfrac{n(n-1)}{2}+\dfrac{(n+1)n}{2}$	M1
$=\dfrac{n\big((n-1)+(n+1)\big)}{2}=\dfrac{2n^{2}}{2}$	A1
$\equiv n^{2}$	AG

Answers: (b) $n=10$

+ Add to test✓ In your test42AASL1-q00051.1 Standard form Paper 1Hard7 marks

Let $x=4\times10^{k}$ and $y=5\times10^{2k-1}$, where $k\in\mathbb{Z}$.

(a)

Find $xy$, giving your answer in the form $a\times10^{m}$, where $1\le a\lt 10$ and $m$ is an expression in $k$.

(2)

(b)

Find $\dfrac{y}{x}$, giving your answer in the form $a\times10^{m}$, where $1\le a\lt 10$ and $m$ is an expression in $k$.

(2)

(c)

Given that $x+y=5.004\times10^{7}$, find the value of $k$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$20\times10^{k+2k-1}=20\times10^{3k-1}$	(M1)
$2\times10^{3k}$	A1
(b)
$\dfrac{5}{4}\times10^{(2k-1)-k}$	(M1)
$1.25\times10^{k-1}$	A1
(c)
recognizing that the leading digit 5 must come from $y$ (so $y$ is the larger term), giving $2k-1=7$	M1	Or a valid trial of values of $k$.
$k=4$	A1
check: $x+y=40\,000+50\,000\,000=50\,040\,000=5.004\times10^{7}$	R1	Must see the check (or equivalent) for R1.

Answers: (a) $2\times10^{3k}$ · (b) $1.25\times10^{k-1}$ · (c) $k=4$

+ Add to test✓ In your test43AASL1-q00121.2 Arithmetic sequences and series Paper 1Hard8 marks

The sum of the first $n$ terms of a sequence is given by $S_{n}=3n^{2}-2n$, for $n\in\mathbb{Z}^{+}$.

(a)

Find $u_{1}$ and $u_{2}$.

(3)

(b)

Show that $u_{n}=6n-5$, and hence explain why the sequence is arithmetic.

(3)

(c)

Find the smallest value of $n$ for which $u_{n}\gt 500$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$u_{1}=S_{1}=1$	A1
$u_{2}=S_{2}-S_{1}$, with $S_{2}=8$	(M1)
$u_{2}=7$	A1
(b)
$u_{n}=S_{n}-S_{n-1}=3n^{2}-2n-\big(3(n-1)^{2}-2(n-1)\big)$	M1	For $n\ge2$; $u_{1}=1$ also fits $6n-5$.
$3(2n-1)-2$, leading to $6n-5$	A1	Must see an intermediate simplified step; $6n-5$ is given.
$u_{n+1}-u_{n}=6$ is constant (independent of $n$), so the sequence is arithmetic	R1
(c)
$6n-5\gt 500\Rightarrow n\gt 84.1\ldots$	(M1)
$n=85$	A1

Answers: (a) $u_{1}=1,\ u_{2}=7$ · (c) $n=85$

+ Add to test✓ In your test44AASL1-q00201.2 Arithmetic sequences and series Paper 1Hard14 marks

An arithmetic sequence has first term 4 and common difference $d$, where $d\neq0$. The first, third and ninth terms of this arithmetic sequence are, in that order, the first three terms of a geometric sequence.

(a)

Show that $d=4$.

(4)

(b)

Find the common ratio of the geometric sequence.

(1)

(c)

Show that the sum of the first $n$ terms of the arithmetic sequence is $2n^{2}+2n$.

(2)

(d)

Find an expression for the sum of the first $n$ terms of the geometric sequence, in the form $a\left(3^{n}-1\right)$, where $a\in\mathbb{Z}$.

(2)

(e)

Find which term of the arithmetic sequence is equal to the fifth term of the geometric sequence.

(3)

(f)

Prove that every term of the geometric sequence is also a term of the arithmetic sequence.

(2)

Show mark scheme

Working	Mark	Notes
(a)
third term $4+2d$, ninth term $4+8d$	A1
$\dfrac{4+2d}{4}=\dfrac{4+8d}{4+2d}$, or $(4+2d)^{2}=4(4+8d)$	M1
$16+16d+4d^{2}=16+32d$, so $4d^{2}-16d=0$	A1
$4d(d-4)=0$ and $d\neq0$	R1	Must reject $d=0$.
$d=4$	AG
(b)
$r=3$	A1	Terms 4, 12, 36.
(c)
$\dfrac{n}{2}\big(2(4)+4(n-1)\big)$	M1
$\dfrac{n}{2}(4n+4)$	A1
$=2n^{2}+2n$	AG
(d)
$\dfrac{4\left(3^{n}-1\right)}{3-1}$	(M1)
$2\left(3^{n}-1\right)$	A1
(e)
fifth term of the geometric sequence $=4\times3^{4}=324$	(A1)
$4+4(n-1)=324$, i.e. $4n=324$	(M1)
the 81st term	A1
(f)
$n$th arithmetic term is $4n$; $k$th geometric term is $4\times3^{k-1}$	M1
$4\times3^{k-1}=4n$ with $n=3^{k-1}$, which is a positive integer for every $k\in\mathbb{Z}^{+}$; so each geometric term is the $\left(3^{k-1}\right)$th arithmetic term	R1

Answers: (b) $r=3$ · (d) $2\left(3^{n}-1\right)$ · (e) the 81st term

+ Add to test✓ In your test45AASL1-q00211.2 Arithmetic sequences and series Paper 1Hard13 marks

Two arithmetic sequences are defined as follows.

Sequence A: $5,\ 12,\ 19,\ 26,\ \ldots$
Sequence B: $3,\ 8,\ 13,\ 18,\ \ldots$

(a)(i)

Find an expression for the $n$th term of sequence A.

(1)

(a)(ii)

Find an expression for the $m$th term of sequence B.

(1)

(b)

Find the sum of the first 50 terms of sequence A.

(2)

(c)

Show that 33 is a term of both sequences.

(2)

(d)

The numbers that are terms of both sequences form a third sequence, C, with first term 33. Explain why C is arithmetic with common difference 35.

(2)

(e)

Find the number of terms of sequence C that are less than 1000.

(3)

(f)

Find the sum of the terms of sequence C that are less than 1000.

(2)

Show mark scheme

Working	Mark	Notes
(a)(i)
$7n-2$	A1
(a)(ii)
$5m-2$	A1
(b)
$\dfrac{50}{2}\big(2(5)+49(7)\big)$	(M1)
$8825$	A1
(c)
$7n-2=33\Rightarrow n=5$, an integer	A1
$5m-2=33\Rightarrow m=7$, an integer	A1
(d)
$7n-2=5m-2\Rightarrow7n=5m$, so $n$ must be a multiple of 5 (as 5 and 7 have no common factor)	R1	Or: a common term must increase by a multiple of both 7 and 5 to reach the next common term.
common terms are $7(5k)-2=35k-2$, so consecutive common terms differ by 35	R1
(e)
$33+35(k-1)\lt 1000$, i.e. $35k-2\lt 1000$	(M1)
$k\lt 28.6\ldots$, or 28th term 978 and 29th term 1013	(A1)
28 terms	A1
(f)
$\dfrac{28}{2}(33+978)$	(M1)	FT from (e).
$14\,154$	A1

Answers: (a) $7n-2$, $5m-2$ · (b) 8825 · (e) 28 · (f) 14 154

+ Add to test✓ In your test46AASL1-q00261.3 Geometric sequences and series Paper 1Hard9 marks

A geometric sequence has third term $18$, and the sum of its first three terms is $26$.

(a)

Show that the common ratio, $r$, satisfies $4r^{2}-9r-9=0$.

(3)

(b)

Find the two possible values of $r$, and the corresponding values of the first term.

(3)

(c)

For the sequence in which $r\gt 0$, find the least value of $n$ for which $S_{n}\gt 10^{6}$. You may use $3^{6}=729$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$u_{1}r^{2}=18$ and $u_{1}(1+r+r^{2})=26$	A1
eliminating $u_{1}$: $\dfrac{1+r+r^{2}}{r^{2}}=\dfrac{26}{18}$	M1
$9+9r+9r^{2}=13r^{2}$	A1
$4r^{2}-9r-9=0$	AG
(b)
$(4r+3)(r-3)=0$, so $r=3$ or $r=-\dfrac{3}{4}$	A1
$r=3\Rightarrow u_{1}=2$	A1
$r=-\dfrac{3}{4}\Rightarrow u_{1}=32$	A1
(c)
$S_{n}=\dfrac{2\left(3^{n}-1\right)}{3-1}=3^{n}-1\gt 10^{6}$	M1
$3^{12}=531\,441$ and $3^{13}=1\,594\,323$	(A1)
$n=13$	A1

Answers: (b) $r=3,\ u_{1}=2$ or $r=-\tfrac{3}{4},\ u_{1}=32$ · (c) $n=13$

+ Add to test✓ In your test47AASL1-q00591.6 Deductive proof Paper 1Hard7 marks

Let $n\in\mathbb{Z}^{+}$.

(a)

Show that $\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\equiv\sqrt{n+1}-\sqrt{n}$.

(2)

(b)

Hence find the value of $\displaystyle\sum_{n=1}^{99}\frac{1}{\sqrt{n+1}+\sqrt{n}}$.

(3)

(c)

Find the least value of $N$ for which $\displaystyle\sum_{n=1}^{N}\frac{1}{\sqrt{n+1}+\sqrt{n}}\gt 20$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\times\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$	M1
denominator $(n+1)-n=1$	A1
$\equiv\sqrt{n+1}-\sqrt{n}$	AG
(b)
$\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\cdots+\left(\sqrt{100}-\sqrt{99}\right)$	M1	At least the first two and the last term.
intermediate terms cancel, leaving $\sqrt{100}-\sqrt{1}$	M1
$9$	A1
(c)
$\sqrt{N+1}-1\gt 20$, so $N+1\gt 441$	(M1)
$N=441$	A1	Award A0 for 440 (the sum is then exactly 20).

Answers: (b) 9 · (c) $N=441$

+ Add to test✓ In your test48AASL1-q00631.7 Rational exponents and laws of logarithms Paper 1Hard7 marks

(a)

Show that $\log_{4}\left(x^{2}\right)=\log_{2}x$, for $x\gt 0$.

(2)

(b)

Hence solve $\log_{2}x-\log_{4}(x+12)=\dfrac{1}{2}$.

(5)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{\log_{2}\left(x^{2}\right)}{\log_{2}4}$	M1
$\dfrac{2\log_{2}x}{2}$	A1
$=\log_{2}x$	AG
(b)
$\log_{4}\left(x^{2}\right)-\log_{4}(x+12)=\tfrac{1}{2}$	M1
$\log_{4}\left(\dfrac{x^{2}}{x+12}\right)=\tfrac{1}{2}$	M1
$\dfrac{x^{2}}{x+12}=4^{\frac{1}{2}}=2$	A1
$x^{2}-2x-24=0$, so $x=6$ or $x=-4$	A1
$x=6$ only	A1	Award the final A1 only if $x=-4$ is rejected (since $\log_{2}x$ requires $x\gt 0$) or not given.

Answers: (b) $x=6$

+ Add to test✓ In your test49AASL1-q00681.7 Rational exponents and laws of logarithms Paper 1Hard7 marks

Let $x\gt 0$, $x\neq1$.

(a)

Show that $\log_{x}2=\dfrac{1}{\log_{2}x}$.

(2)

(b)

Hence solve $\log_{2}x+3\log_{x}2=4$.

(5)

Show mark scheme

Working	Mark	Notes
(a)
$\log_{x}2=\dfrac{\log_{2}2}{\log_{2}x}$	M1	Change of base to base 2 (or to any common base).
$\log_{2}2=1$	A1
$\log_{x}2=\dfrac{1}{\log_{2}x}$	AG
(b)
$\log_{2}x+\dfrac{3}{\log_{2}x}=4$	M1
letting $y=\log_{2}x$: $y^{2}-4y+3=0$	M1
$y=1$ or $y=3$	A1
$x=2^{1}$ or $x=2^{3}$	(M1)
$x=2$ or $x=8$	A1	Both values required.

Answers: (b) $x=2$ or $x=8$

+ Add to test✓ In your test50AASL1-q00701.7 Rational exponents and laws of logarithms Paper 1Hard10 marks

Consider the simultaneous equations

$2^{x}\times4^{y}=64$ and $\log_{2}x+\log_{2}y=2$, where $x,y\gt 0$.

(a)

Show that $x+2y=6$.

(2)

(b)

Show that $xy=4$.

(2)

(c)

Hence find all pairs $(x,y)$ that satisfy both equations.

(4)

(d)

The second equation is replaced by $\log_{2}x+\log_{2}y=3$. Explain why the new pair of equations has no solution.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$4^{y}=2^{2y}$	M1
$2^{x+2y}=2^{6}$	A1
$x+2y=6$	AG
(b)
$\log_{2}(xy)=2$	M1
$xy=2^{2}$	A1
$xy=4$	AG
(c)
substituting, e.g. $(6-2y)y=4$	M1
$2y^{2}-6y+4=0$, i.e. $y^{2}-3y+2=0$	A1	Or $x^{2}-6x+8=0$.
$y=1$ or $y=2$	A1
$(4,\,1)$ and $(2,\,2)$	A1	Both pairs required.
(d)
$xy=8$ leads to $(6-2y)y=8$, i.e. $y^{2}-3y+4=0$	M1
discriminant $=9-16=-7\lt 0$, so there are no real solutions	R1

Answers: (c) $(4,\,1)$ and $(2,\,2)$

+ Add to test✓ In your test51AASL1-q00721.7 Rational exponents and laws of logarithms Paper 1Hard16 marks

A geometric sequence has first term $u_{1}=2$ and common ratio $8$. A second sequence is defined by $v_{n}=\log_{2}u_{n}$, for $n\in\mathbb{Z}^{+}$.

(a)

Write down the values of $u_{2}$ and $u_{3}$.

(1)

(b)

Show that $v_{n}=3n-2$.

(3)

(c)

Hence explain why the sequence $v_{1},v_{2},v_{3},\ldots$ is arithmetic, and state its common difference.

(2)

(d)

Find $\displaystyle\sum_{n=1}^{20}v_{n}$.

(2)

(e)

Hence find the value of $\log_{2}\left(u_{1}u_{2}u_{3}\cdots u_{20}\right)$.

(2)

(f)

Prove that, for any geometric sequence with first term $a\gt 0$ and common ratio $r\gt 0$, the sequence $\ln u_{1},\ln u_{2},\ln u_{3},\ldots$ is arithmetic.

(2)

(g)

A geometric sequence $w_{1},w_{2},w_{3},\ldots$ has positive terms, with $\log_{10}w_{1}=5$ and $\log_{10}w_{6}=-5$. Find the sum to infinity of this sequence.

(4)

Show mark scheme

Working	Mark	Notes
(a)
$u_{2}=16,\ u_{3}=128$	A1	Both required.
(b)
$u_{n}=2\times8^{n-1}$	M1
$v_{n}=\log_{2}2+\log_{2}8^{n-1}=\log_{2}2+(n-1)\log_{2}8$	M1
$=1+3(n-1)$	A1
$v_{n}=3n-2$	AG
(c)
$v_{n+1}-v_{n}=3(n+1)-2-(3n-2)=3$, which is independent of $n$	R1
$d=3$	A1
(d)
$\dfrac{20}{2}\big(v_{1}+v_{20}\big)=10(1+58)$	(M1)
$590$	A1
(e)
$\log_{2}\left(u_{1}u_{2}\cdots u_{20}\right)=\log_{2}u_{1}+\log_{2}u_{2}+\cdots+\log_{2}u_{20}=\displaystyle\sum_{n=1}^{20}v_{n}$	M1
$590$	A1	FT from (d).
(f)
$\ln u_{n}=\ln\left(ar^{n-1}\right)=\ln a+(n-1)\ln r$	M1
$\ln u_{n+1}-\ln u_{n}=\ln r$, a constant (independent of $n$), so the sequence is arithmetic	R1
(g)
the logarithms form an arithmetic sequence with $5+5d=-5$, so $d=-2$	M1	Or $\dfrac{w_{6}}{w_{1}}=\dfrac{10^{-5}}{10^{5}}=r^{5}$.
$r=10^{-2}=\tfrac{1}{100}$	A1
$S_{\infty}=\dfrac{10^{5}}{1-\frac{1}{100}}$	M1	FT from their r, provided \|r\| < 1.
$\dfrac{10^{7}}{99}$	A1	Accept $101\,010.\dot{1}\dot{0}$.

Answers: (a) 16, 128 · (c) $d=3$ · (d) 590 · (e) 590 · (g) $\dfrac{10^{7}}{99}$

+ Add to test✓ In your test52AASL1-q00731.8 Infinite geometric series Paper 1Hard8 marks

Consider the infinite geometric series $1+(2x-1)+(2x-1)^{2}+(2x-1)^{3}+\cdots$.

(a)

Find the set of values of $x$ for which the series converges.

(2)

(b)

Show that, for these values of $x$, the sum to infinity is $\dfrac{1}{2-2x}$.

(2)

(c)

Find the value of $x$ for which the sum to infinity is $\dfrac{4}{3}$.

(2)

(d)

Explain why the sum to infinity cannot be equal to $\dfrac{1}{3}$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\|2x-1\|\lt 1$	(M1)
$0\lt x\lt 1$	A1
(b)
$S_{\infty}=\dfrac{1}{1-(2x-1)}$	M1
denominator simplified: $1-2x+1$	A1
$=\dfrac{1}{2-2x}$	AG
(c)
$2-2x=\dfrac{3}{4}$	(M1)
$x=\dfrac{5}{8}$	A1
(d)
solving $\dfrac{1}{2-2x}=\dfrac{1}{3}$, giving $x=-\tfrac{1}{2}$; or noting $0\lt 2-2x\lt 2$ for $0\lt x\lt 1$	M1
$x=-\tfrac{1}{2}$ is not in $0\lt x\lt 1$, so the series does not converge there; or $S_{\infty}\gt \tfrac{1}{2}$ for all valid $x$	R1

Answers: (a) $0\lt x\lt 1$ · (c) $x=\tfrac{5}{8}$

+ Add to test✓ In your test53AASL1-q00761.8 Infinite geometric series Paper 1Hard10 marks

An infinite geometric series has sum to infinity 9 and second term 2.

(a)

Show that the common ratio, $r$, satisfies $9r^{2}-9r+2=0$.

(3)

(b)

Find the two possible values of $r$.

(2)

(c)

Find the first term corresponding to each value of $r$.

(2)

(d)

Explain why both values of $r$ give a series whose sum to infinity exists.

(1)

(e)

Find the positive difference between the sums of the first three terms of the two series.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{u_{1}}{1-r}=9$, so $u_{1}=9(1-r)$	M1
$u_{1}r=2$, so $9r(1-r)=2$	M1
$9r-9r^{2}=2$	A1
$9r^{2}-9r+2=0$	AG
(b)
$(3r-1)(3r-2)=0$	(M1)
$r=\tfrac{1}{3}$ or $r=\tfrac{2}{3}$	A1
(c)
$r=\tfrac{1}{3}$: $u_{1}=6$	A1
$r=\tfrac{2}{3}$: $u_{1}=3$	A1
(d)
$\left\|\tfrac{1}{3}\right\|\lt 1$ and $\left\|\tfrac{2}{3}\right\|\lt 1$	R1
(e)
$6+2+\tfrac{2}{3}=\tfrac{26}{3}$ and $3+2+\tfrac{4}{3}=\tfrac{19}{3}$	(M1)	Award for either sum correct.
$\tfrac{7}{3}$	A1

Answers: (b) $\tfrac{1}{3}$ or $\tfrac{2}{3}$ · (c) 6 and 3 · (e) $\tfrac{7}{3}$

+ Add to test✓ In your test54AASL1-q00791.8 Infinite geometric series Paper 1Hard15 marks

Square $S_{1}$ has sides of length 8 cm. Square $S_{2}$ is formed by joining the midpoints of the sides of $S_{1}$. Square $S_{3}$ is formed by joining the midpoints of the sides of $S_{2}$, and the process continues indefinitely, as shown in the diagram.

(a)

Show that the length of a side of $S_{2}$ is $4\sqrt{2}$ cm.

(2)

(b)

Show that the areas of $S_{1},S_{2},S_{3},\ldots$ form a geometric sequence with common ratio $\tfrac{1}{2}$.

(2)

(c)

Find the area of $S_{6}$.

(2)

(d)

Find the sum to infinity of the areas of the squares.

(2)

(e)

The perimeters of $S_{1},S_{2},S_{3},\ldots$ also form a geometric sequence. Find the sum to infinity of the perimeters, giving your answer in the form $a+b\sqrt{2}$, where $a,b\in\mathbb{Z}$.

(4)

(f)

Find the least value of $n$ for which the perimeter of $S_{n}$ is less than 1 cm.

(3)

Show mark scheme

Working	Mark	Notes
(a)
each half-side of $S_{1}$ is 4 cm; using Pythagoras: $\sqrt{4^{2}+4^{2}}$	M1
$\sqrt{32}=\sqrt{16\times2}$	A1
$=4\sqrt{2}$ cm	AG
(b)
each side is $\dfrac{1}{\sqrt{2}}$ times the previous side (by the same argument as part (a))	M1	Or $\dfrac{4\sqrt{2}}{8}=\dfrac{1}{\sqrt{2}}$ applied generally.
so each area is $\left(\dfrac{1}{\sqrt{2}}\right)^{2}=\dfrac{1}{2}$ times the previous area: $64,\,32,\,16,\ldots$	A1
geometric with common ratio $\tfrac{1}{2}$	AG
(c)
$64\left(\tfrac{1}{2}\right)^{5}$	(M1)
$2$ cm²	A1
(d)
$\dfrac{64}{1-\frac{1}{2}}$	(M1)
$128$ cm²	A1
(e)
first term 32, common ratio $\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$	A1
$\dfrac{32}{1-\frac{1}{\sqrt{2}}}=\dfrac{32\sqrt{2}}{\sqrt{2}-1}$	M1
multiplying by $\dfrac{\sqrt{2}+1}{\sqrt{2}+1}$: $\dfrac{32\sqrt{2}\left(\sqrt{2}+1\right)}{2-1}$	M1
$64+32\sqrt{2}$ cm	A1
(f)
$32\left(\dfrac{1}{\sqrt{2}}\right)^{n-1}\lt 1$, i.e. $\left(\sqrt{2}\right)^{n-1}\gt 32$	M1
$32=2^{5}=\left(\sqrt{2}\right)^{10}$, so $n-1\gt 10$	(M1)	Or perimeter of $S_{11}$ is exactly 1 cm.
$n=12$	A1

Answers: (c) 2 cm² · (d) 128 cm² · (e) $64+32\sqrt{2}$ cm · (f) $n=12$

+ Add to test✓ In your test55AASL1-q00811.9 The binomial theorem Paper 1Hard6 marks

Consider the expansion of $\left(x^{2}-\dfrac{2}{x}\right)^{6}$, where $x\neq0$.

(a)

Show that the general term can be written as $\dbinom{6}{r}(-2)^{r}x^{12-3r}$.

(2)

(b)

Find the constant term.

(3)

(c)

Explain why the expansion has no term in $x^{5}$.

(1)

Show mark scheme

Working	Mark	Notes
(a)
$\dbinom{6}{r}\left(x^{2}\right)^{6-r}\left(-\dfrac{2}{x}\right)^{r}$	M1
$\dbinom{6}{r}(-2)^{r}x^{12-2r}x^{-r}$	A1
$=\dbinom{6}{r}(-2)^{r}x^{12-3r}$	AG
(b)
$12-3r=0$	(M1)
$r=4$	(A1)
$\dbinom{6}{4}(-2)^{4}=15\times16=240$	A1
(c)
$12-3r=5$ gives $r=\tfrac{7}{3}$, which is not an integer	R1

Answers: (b) 240

+ Add to test✓ In your test56AASL1-q00821.9 The binomial theorem Paper 1Hard6 marks

In the expansion of $(1+kx)^{n}$, where $n\in\mathbb{Z}^{+}$ and $k\gt 0$, the coefficient of $x$ is 12 and the coefficient of $x^{2}$ is 60.

(a)

Show that $(n-1)k=10$.

(3)

(b)

Find the value of $n$ and the value of $k$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$nk=12$	A1
$\dbinom{n}{2}k^{2}=\dfrac{n(n-1)}{2}k^{2}=60$	M1
$\dfrac{(nk)(n-1)k}{2}=60\Rightarrow 6(n-1)k=60$	A1	Must see $nk=12$ substituted.
$(n-1)k=10$	AG
(b)
$nk-k=10\Rightarrow 12-k=10$	(M1)
$k=2$	A1
$n=6$	A1

Answers: (b) $n=6,\ k=2$

+ Add to test✓ In your test57AASL1-q00851.9 The binomial theorem Paper 1Hard7 marks

(a)

Use the binomial theorem to show that $\left(1+\sqrt{2}\right)^{5}=41+29\sqrt{2}$.

(3)

(b)

Write down the value of $\left(1-\sqrt{2}\right)^{5}$ in the form $a+b\sqrt{2}$, where $a,b\in\mathbb{Z}$.

(1)

(c)

Given that $0\lt\sqrt{2}-1\lt 1$, hence find the integer $N$ such that $N\lt\left(1+\sqrt{2}\right)^{5}\lt N+1$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$1+5\sqrt{2}+10\left(\sqrt{2}\right)^{2}+10\left(\sqrt{2}\right)^{3}+5\left(\sqrt{2}\right)^{4}+\left(\sqrt{2}\right)^{5}$	M1	Coefficients 1, 5, 10, 10, 5, 1 seen with powers of √2.
$1+5\sqrt{2}+20+20\sqrt{2}+20+4\sqrt{2}$	A1
collecting: $(1+20+20)+(5+20+4)\sqrt{2}$	A1
$=41+29\sqrt{2}$	AG
(b)
$41-29\sqrt{2}$	A1
(c)
$\left(1+\sqrt{2}\right)^{5}+\left(1-\sqrt{2}\right)^{5}=82$	M1
$\left(1-\sqrt{2}\right)^{5}=-\left(\sqrt{2}-1\right)^{5}$, and $0\lt\left(\sqrt{2}-1\right)^{5}\lt 1$, so $\left(1+\sqrt{2}\right)^{5}=82+\left(\sqrt{2}-1\right)^{5}$	R1
$N=82$	A1

Answers: (b) $41-29\sqrt{2}$ · (c) $N=82$

+ Add to test✓ In your test58AASL1-q00061.1 Standard form Paper 2Easy5 marks

One light year is $9.46\times10^{15}$ m. The distance from the Sun to a particular star is $4.0\times10^{16}$ m.

(a)

Find this distance in light years.

(2)

(b)

A space probe travels at a constant speed of $1.7\times10^{4}$ m s⁻¹. Taking one year to be $3.15\times10^{7}$ s, find the time, in years, that the probe would take to travel this distance. Give your answer in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{4.0\times10^{16}}{9.46\times10^{15}}$	(M1)
$4.23$ light years	A1	$4.22832\ldots$
(b)
time $=\dfrac{4.0\times10^{16}}{1.7\times10^{4}}\;(=2.35\times10^{12}\text{ s})$	(M1)
dividing their time in seconds by $3.15\times10^{7}$	(M1)
$7.47\times10^{4}$ years	A1	$74\,696.5\ldots$ Accept $7.46\times10^{4}$ from premature rounding. Award A0 for 74 700 or 7.47E4 (not in the required form).

Answers: (a) 4.23 light years · (b) $7.47\times10^{4}$ years

+ Add to test✓ In your test59AASL1-q00171.2 Arithmetic sequences and series Paper 2Easy5 marks

A finite arithmetic sequence is $12.5,\ 15.2,\ 17.9,\ \ldots,\ 282.5$.

(a)

Write down the common difference.

(1)

(b)

Find the number of terms in the sequence.

(2)

(c)

Find the sum of all the terms of the sequence.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$d=2.7$	A1
(b)
$12.5+(n-1)(2.7)=282.5$	(M1)
$n=101$	A1
(c)
$\dfrac{101}{2}(12.5+282.5)$ or $\dfrac{101}{2}\big(2(12.5)+100(2.7)\big)$	(M1)	FT from (b).
$14\,897.5$	A1	Accept 14 900.

Answers: (a) 2.7 · (b) 101 · (c) 14 897.5

+ Add to test✓ In your test60AASL1-q00281.3 Geometric sequences and series Paper 2Easy5 marks

After a clean-up programme, the concentration of a pollutant in a lake decreases by 30% each year. At the start of the programme the concentration is 2.4 mg per litre.

(a)

Find the concentration after 5 years.

(2)

(b)

Find the least number of complete years needed for the concentration to fall below 0.05 mg per litre.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$2.4\times0.7^{5}$	(M1)	Award M0 for $0.3^{5}$.
$0.403$ mg per litre	A1	$0.403368$
(b)
$2.4\times0.7^{n}\lt 0.05$	(M1)
$n\gt 10.85\ldots$, or after 10 years $0.0678$; after 11 years $0.0475$	(A1)
11 years	A1

Answers: (a) 0.403 mg per litre · (b) 11 years

+ Add to test✓ In your test61AASL1-q00361.4 Financial applications Paper 2Easy5 marks

A car is bought for £28 500. Its value depreciates by 14% each year.

(a)

Find the value of the car after 5 years.

(2)

(b)

Find the number of complete years after which the value of the car first falls below £10 000.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$28\,500\times0.86^{5}$	(M1)
£13 407.17	A1	Accept £13 400.
(b)
$28\,500\times0.86^{n}\lt 10\,000$	(M1)
$n\gt 6.94$, or values £11 530.17 after 6 years and £9915.94 after 7 years	(A1)
7 years	A1

Answers: (a) £13 407.17 · (b) 7 years

+ Add to test✓ In your test62AASL1-q00381.4 Financial applications Paper 2Easy5 marks

Kenji invests ¥800 000 in an account that pays a nominal annual interest rate of 3.6%, compounded half-yearly.

(a)

Find the value of the investment after 7 years. Give your answer to the nearest yen.

(2)

(b)

Find the interest earned in these 7 years.

(1)

(c)

Find the least number of years, given as a multiple of 0.5, after which the value of the investment is at least ¥1 000 000.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$800\,000\left(1+\dfrac{3.6}{200}\right)^{14}$, or TVM with $N=14,\ I\%=3.6,\ PV=-800\,000,\ P/Y=2,\ C/Y=2$	(M1)
¥1 026 973	A1	$1\,026\,972.65\ldots$
(b)
¥226 973	A1	FT from (a).
(c)
$800\,000(1.018)^{n}\ge1\,000\,000\Rightarrow n\ge12.5\ldots$ half-years; or 12 half-years: ¥990 976; 13 half-years: ¥1 008 814	(M1)
6.5 years	A1	Award A0 for 13 without converting to years.

Answers: (a) ¥1 026 973 · (b) ¥226 973 · (c) 6.5 years

+ Add to test✓ In your test63AASL1-q00501.5 Exponents and introduction to logarithms Paper 2Easy4 marks

The sound intensity level, $L$ decibels (dB), of a sound of intensity $I$ W m⁻² is given by $L=10\log_{10}\left(\dfrac{I}{I_{0}}\right)$, where $I_{0}=10^{-12}$ W m⁻².

(a)

Find $L$ for a sound of intensity $3.2\times10^{-5}$ W m⁻².

(2)

(b)

Find the intensity of a sound with a level of 85 dB.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$10\log_{10}\left(\dfrac{3.2\times10^{-5}}{10^{-12}}\right)$	(M1)
$75.1$ dB	A1	75.0514…
(b)
$\log_{10}\left(\dfrac{I}{10^{-12}}\right)=8.5$, so $I=10^{-12}\times10^{8.5}$	(M1)
$3.16\times10^{-4}$ W m⁻²	A1	Accept 0.000316.

Answers: (a) 75.1 dB · (b) $3.16\times10^{-4}$ W m⁻²

+ Add to test✓ In your test64AASL1-q00071.1 Standard form Paper 2Medium6 marks

A red blood cell is modelled as a sphere of radius $2.8\times10^{-6}$ m.

(a)

Find the volume of one red blood cell, according to the model. Give your answer in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(2)

(b)

A blood sample has a volume of $5.0\times10^{-6}$ m³. Red blood cells make up 45% of the volume of the sample. Use the model to find the number of red blood cells in the sample. Give your answer in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(2)

(c)

An adult has $2.5\times10^{13}$ red blood cells. The body produces new red blood cells at a rate of $2.4\times10^{6}$ cells per second. Find the number of days the body takes to produce $2.5\times10^{13}$ cells.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{4}{3}\pi\left(2.8\times10^{-6}\right)^{3}$	(M1)
$9.20\times10^{-17}$ m³	A1	$9.1952\ldots\times10^{-17}$
(b)
$\dfrac{0.45\times5.0\times10^{-6}}{9.1952\times10^{-17}}$	(M1)	FT from (a).
$2.45\times10^{10}$	A1	$2.4469\ldots\times10^{10}$. Accept $2.44\times10^{10}$–$2.45\times10^{10}$. Award A0 for 2.45E10.
(c)
$\dfrac{2.5\times10^{13}}{2.4\times10^{6}}=1.0417\times10^{7}$ s, divided by $86\,400$	(M1)	Accept $24\times60\times60$.
$121$ days	A1	$120.56\ldots$ Accept 120.6 or 120.

Answers: (a) $9.20\times10^{-17}$ m³ · (b) $2.45\times10^{10}$ · (c) 121 days

+ Add to test✓ In your test65AASL1-q00161.2 Arithmetic sequences and series Paper 2Medium6 marks

The table shows the number of members of a cycling club at the start of each year from 2019 to 2023.

Year	2019	2020	2021	2022	2023
Members	412	437	465	488	514

(a)

Show that the numbers of members do not form an exact arithmetic sequence.

(1)

(b)

Using the values for 2019 and 2023, find an estimate for the common difference.

(2)

(c)

The club models the number of members with an arithmetic sequence, with $u_{1}=412$ in 2019 and the common difference found in part (b). Use the model to predict the number of members in 2030.

(2)

(d)

Comment on the reliability of the prediction in part (c).

(1)

Show mark scheme

Working	Mark	Notes
(a)
differences are 25, 28, 23, 26, which are not all equal	A1	Must see at least two different differences.
(b)
$\dfrac{514-412}{4}$	(M1)
$25.5$	A1
(c)
$u_{12}=412+11(25.5)$	(M1)	FT from (b). Award M0 for $u_{11}$ or $10d$.
$692.5$, so about 692 or 693 members	A1	Accept 692, 692.5 or 693.
(d)
not reliable: 2030 is far outside the range of the data (extrapolation), and the differences are not constant	R1	Accept any valid comment that refers to extrapolation or to the varying differences.

Answers: (b) 25.5 · (c) about 693 members

+ Add to test✓ In your test66AASL1-q00181.2 Arithmetic sequences and series Paper 2Medium8 marks

A cyclist follows a training plan. She rides 18 km on day 1, and on each following day she rides 2 km more than on the previous day.

(a)

Find the distance she rides on day 25.

(2)

(b)

Find the least number of days needed for the total distance ridden to be at least 1500 km.

(3)

(c)

The plan is changed so that, once her daily distance reaches 50 km, she rides exactly 50 km each day after that. Find the total distance she rides in the first 30 days under the changed plan.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$18+24(2)$	(M1)
$66$ km	A1
(b)
$\dfrac{n}{2}\big(36+2(n-1)\big)\ge1500$, i.e. $n^{2}+17n-1500\ge0$	(M1)
$n\ge31.2$, or $S_{31}=1488$ and $S_{32}=1568$	(A1)
32 days	A1
(c)
$18+2(n-1)=50\Rightarrow$ day 17	(M1)
$\dfrac{17}{2}(18+50)+13\times50$	(M1)	$578+650$
$1228$ km	A1

Answers: (a) 66 km · (b) 32 days · (c) 1228 km

+ Add to test✓ In your test67AASL1-q00191.2 Arithmetic sequences and series Paper 2Medium6 marks

Two friends start saving in the same week.

Leila saves $2.50 in week 1 and increases the amount she saves by $0.40 each week.
Tomás saves $5.00 in week 1 and increases the amount he saves by $0.25 each week.

(a)

Find the first week in which Leila saves more than Tomás in that week.

(3)

(b)

Find the first week at the end of which Leila’s total savings are greater than Tomás’s total savings.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$2.5+0.4(n-1)\gt 5+0.25(n-1)$	(M1)
$n-1\gt 16.7\ldots$, or week 17: $8.90 and $9.00; week 18: $9.30 and $9.25	(A1)
week 18	A1
(b)
$\dfrac{n}{2}\big(5+0.4(n-1)\big)\gt \dfrac{n}{2}\big(10+0.25(n-1)\big)$	(M1)
$n-1\gt 33.3\ldots$, or week 34: $309.40 and $310.25; week 35: $325.50 and $323.75	(A1)
week 35	A1

Answers: (a) week 18 · (b) week 35

+ Add to test✓ In your test68AASL1-q00271.3 Geometric sequences and series Paper 2Medium7 marks

In the first month after its launch, an app had 2400 new downloads. In each following month, the number of new downloads was 15% greater than in the previous month.

(a)

Find the number of new downloads in the sixth month.

(2)

(b)(i)

The total number of downloads in the first 12 months.

Write down an expression, using sigma notation, for this total.

(1)

(b)(ii)

Find this total.

(2)

(c)

Find the month in which the total number of downloads first exceeds 200 000.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$2400\times1.15^{5}$	(M1)
$4827$	A1	$4827.26\ldots$ Accept 4830.
(b)(i)
$\displaystyle\sum_{n=1}^{12}2400(1.15)^{n-1}$	A1	Accept equivalent forms, e.g. $\sum_{n=0}^{11}2400(1.15)^{n}$.
(b)(ii)
$\dfrac{2400(1.15^{12}-1)}{1.15-1}$	(M1)
$69\,604$	A1	$69\,604.0\ldots$ Accept 69 600.
(c)
$\dfrac{2400(1.15^{n}-1)}{0.15}\gt 200\,000$, or table: $S_{18}=182\,007$, $S_{19}=211\,708$	(M1)	$n\gt 18.6$
the 19th month	A1

Answers: (a) 4827 · (b)(ii) 69 604 · (c) 19th month

+ Add to test✓ In your test69AASL1-q00291.3 Geometric sequences and series Paper 2Medium8 marks

The population of a town was 24 500 at the start of 2015 and 28 100 at the start of 2020. The population at the start of each year is modelled by a geometric sequence.

(a)

Find the common ratio of the sequence, giving your answer correct to four significant figures.

(3)

(b)

Use the model to estimate the population at the start of 2027.

(2)

(c)

Find the first year in which the population at the start of the year exceeds 40 000, according to the model.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$24\,500r^{5}=28\,100$	(M1)	Award M0 for $r^{4}$ or $r^{6}$.
$r=\left(\dfrac{28\,100}{24\,500}\right)^{\frac{1}{5}}$	(A1)
$r=1.028$	A1	$1.02779\ldots$
(b)
$24\,500r^{12}$	(M1)
$34\,000$	A1	$34\,045.7\ldots$ with unrounded $r$; $34\,126$ with $r=1.028$. Accept answers from 34 000 to 34 130.
(c)
$24\,500r^{n}\gt 40\,000$	(M1)
$n\gt 17.9$ (or 17.8 with $r=1.028$); or start of 2032: 39 048, start of 2033: 40 134	(A1)
2033	A1

+ Add to test✓ In your test70AASL1-q00301.3 Geometric sequences and series Paper 2Medium9 marks

The table shows the number of active users of a new online game in each of its first five weeks.

Week	1	2	3	4	5
Active users	1200	1440	1730	2070	2490

(a)

Show that the numbers of users are approximately, but not exactly, a geometric sequence.

(2)

(b)

Use the values for weeks 1 and 5 to estimate the common ratio.

(2)

(c)

The numbers of users are modelled by a geometric sequence with first term 1200 and common ratio 1.2. Use this model to find the number of users in week 10.

(2)

(d)

Use the model in part (c) to find the first week in which the number of users exceeds 20 000.

(2)

(e)

Suggest one reason why the model may not be valid for predicting the number of users after several months.

(1)

Show mark scheme

Working	Mark	Notes
(a)
finding the ratios of consecutive terms: $1.2,\ 1.201,\ 1.197,\ 1.203$	M1	At least three ratios.
the ratios are all close to (but not exactly equal to) 1.2	R1
(b)
$1200r^{4}=2490$	(M1)
$r=1.20$	A1	$1.2002\ldots$
(c)
$1200(1.2)^{9}$	(M1)
$6192$	A1	$6191.7\ldots$ Accept 6190.
(d)
$1200(1.2)^{n-1}\gt 20\,000$, or week 16: 18 488; week 17: 22 186	(M1)
week 17	A1
(e)
e.g. the number of potential players is limited, so growth cannot continue at 20% per week indefinitely; interest in the game may fall	R1	Accept any reasonable comment about extrapolation or limits on growth.

Answers: (b) 1.20 · (c) 6192 · (d) week 17

+ Add to test✓ In your test71AASL1-q00311.3 Geometric sequences and series Paper 2Medium9 marks

In its first year, a mine produces 52 000 tonnes of ore. In each following year, the output is 6% less than in the previous year. The mine closes at the end of the last year in which its output is at least 20 000 tonnes.

(a)

Find the output of the mine in its 8th year.

(2)

(b)

Find the total output of the mine in its first 8 years.

(2)

(c)

Find the number of years for which the mine operates.

(3)

(d)

Find the total output of the mine over this time.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$52\,000(0.94)^{7}$	(M1)
$33\,700$ tonnes	A1	$33\,720.8\ldots$
(b)
$\dfrac{52\,000\left(1-0.94^{8}\right)}{1-0.94}$	(M1)
$338\,000$ tonnes	A1	$338\,373.6\ldots$
(c)
$52\,000(0.94)^{n-1}\ge20\,000$	(M1)
$n\le16.4$; or year 16: $20\,555$; year 17: $19\,322$	(A1)
16 years	A1
(d)
$\dfrac{52\,000\left(1-0.94^{16}\right)}{0.06}$	(M1)	FT from (c).
$545\,000$ tonnes	A1	$544\,635.6\ldots$

Answers: (a) 33 700 t · (b) 338 000 t · (c) 16 years · (d) 545 000 t

+ Add to test✓ In your test72AASL1-q00351.4 Financial applications Paper 2Medium6 marks

Ana invests €5000 in an account that pays simple interest at a rate of 4.5% per year. Ben invests €5000 in an account that pays a nominal annual interest rate of 3.8%, compounded monthly.

(a)

Find the value of Ana’s investment after 10 years.

(2)

(b)

Find the value of Ben’s investment after 10 years. Give your answer to the nearest cent.

(2)

(c)

Find the number of complete years after which the value of Ben’s investment first exceeds the value of Ana’s investment.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$5000+10(0.045\times5000)$ (arithmetic sequence with $d=225$)	(M1)
€7250	A1
(b)
$5000\left(1+\dfrac{3.8}{1200}\right)^{120}$, or TVM with $N=120,\ I\%=3.8,\ PV=-5000,\ P/Y=12,\ C/Y=12$	(M1)
€7307.03	A1
(c)
comparing $5000+225n$ with $5000\left(1+\frac{3.8}{1200}\right)^{12n}$, e.g. table or graph	(M1)	e.g. after 8 years: 6800 and 6773.09; after 9 years: 7025 and 7035.00
9 years	A1

Answers: (a) €7250 · (b) €7307.03 · (c) 9 years

+ Add to test✓ In your test73AASL1-q00371.4 Financial applications Paper 2Medium7 marks

Sofia invests $12 000 in an account that pays a nominal annual interest rate of 5.2%, compounded quarterly.

(a)

Find the value of the investment after 8 years. Give your answer to the nearest dollar.

(2)

(b)

Find the minimum number of complete years needed for the investment to at least double in value.

(2)

(c)

The annual rate of inflation is 2.5% for the 8 years. Find the real value of the investment after 8 years, in terms of today’s money. Give your answer to the nearest dollar.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$12\,000\left(1+\dfrac{5.2}{400}\right)^{4\times8}$, or TVM with $N=32,\ I\%=5.2,\ PV=-12\,000,\ P/Y=4,\ C/Y=4$	(M1)
$18 142	A1	$18 141.93…
(b)
$12\,000\left(1+\dfrac{5.2}{400}\right)^{4n}\ge 24\,000$, or TVM giving $N=53.7$ quarters (13.4 years)	(M1)
14 years	A1
(c)
attempt to adjust for inflation, e.g. dividing by $1.025^{n}$ or using $5.2-2.5$	(M1)
$\dfrac{18\,141.93}{1.025^{8}}$	(A1)
$14 890	A1	$14 889.93… Also accept the real-interest-rate method, $12\,000\left(1+\frac{5.2-2.5}{400}\right)^{32}=14\,882$. Accept answers between $14 882 and $14 890.

Answers: (a) $18 142 · (b) 14 years · (c) $14 890

+ Add to test✓ In your test74AASL1-q00391.4 Financial applications Paper 2Medium6 marks

An investment of £15 000 grows to £20 000 in exactly 6 years, with interest compounded monthly at a fixed nominal annual rate of $r\%$.

(a)

Find the value of $r$.

(2)

(b)

Find the annual interest rate, compounded annually, that would give the same growth over the 6 years.

(2)

(c)

The investment remains in the account at the same rate $r\%$, compounded monthly. Find its value after a total of 10 years.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$15\,000\left(1+\dfrac{r}{1200}\right)^{72}=20\,000$, or TVM with $N=72,\ PV=-15\,000,\ FV=20\,000,\ P/Y=12,\ C/Y=12$	(M1)
$r=4.80$	A1	$4.8042\ldots$
(b)
$15\,000\left(1+\dfrac{R}{100}\right)^{6}=20\,000$	(M1)
$4.91\%$	A1	$4.9115\ldots$
(c)
$15\,000\left(1+\dfrac{4.8042}{1200}\right)^{120}$	(M1)
£24 228	A1	$24\,228.27\ldots$ Accept £24 218 (from $r=4.80$) to £24 229.

Answers: (a) $r=4.80$ · (b) 4.91% · (c) £24 228

+ Add to test✓ In your test75AASL1-q00401.4 Financial applications Paper 2Medium5 marks

Lena needs $50 000 in 10 years’ time. She can choose between two accounts.

Account A: nominal annual interest rate 4.2%, compounded quarterly.
Account B: interest rate 4.2% per year, compounded annually.

(a)

Find the amount Lena must invest now in Account A. Give your answer to the nearest cent.

(2)

(b)

Find the amount Lena must invest now in Account B. Give your answer to the nearest cent.

(2)

(c)

Explain why the amount in part (a) is smaller than the amount in part (b).

(1)

Show mark scheme

Working	Mark	Notes
(a)
$PV\left(1+\dfrac{4.2}{400}\right)^{40}=50\,000$, or TVM with $N=40,\ I\%=4.2,\ FV=50\,000,\ P/Y=4,\ C/Y=4$	(M1)
$32 924.36	A1
(b)
$PV(1.042)^{10}=50\,000$	(M1)
$33 135.45	A1
(c)
with quarterly compounding, interest is added more often and itself earns interest sooner, so the same nominal rate gives greater growth	R1	Accept equivalent: the effective annual rate of A (4.27%) is greater than 4.2%.

Answers: (a) $32 924.36 · (b) $33 135.45

+ Add to test✓ In your test76AASL1-q00411.4 Financial applications Paper 2Medium8 marks

A piece of factory equipment was bought for $64 000. It depreciates by a fixed percentage each year. After 4 years it is valued at $31 000.

(a)

Find the annual rate of depreciation, giving your answer as a percentage correct to three significant figures.

(3)

(b)

Find the value of the equipment after 8 years.

(2)

(c)

The equipment is replaced at the end of the first year in which its value falls below 10% of its purchase price. Find the number of years for which the equipment is used.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$64\,000\,x^{4}=31\,000$, where $x$ is the annual multiplier	(M1)
$x=0.83424\ldots$	(A1)
$16.6\%$	A1	$16.575\ldots\%$
(b)
$64\,000(0.83424\ldots)^{8}$, or $31\,000\times\dfrac{31\,000}{64\,000}$	(M1)
$15 016	A1	$15\,015.625$. Accept $14 980 (from 16.6%) to $15 016.
(c)
$64\,000(0.83424\ldots)^{n}\lt 6400$	(M1)
$n\gt 12.7$; or after 12 years $7273, after 13 years $6068	(A1)
13 years	A1

Answers: (a) 16.6% · (b) $15 016 · (c) 13 years

+ Add to test✓ In your test77AASL1-q00511.5 Exponents and introduction to logarithms Paper 2Medium8 marks

Let $N=7^{200}$. The number $N$ is too large to be evaluated directly on a calculator.

(a)

Find the value of $\log_{10}N$, giving your answer correct to three decimal places.

(2)

(b)

Hence write $N$ in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$, giving the value of $a$ correct to three significant figures.

(3)

(c)

Write down the number of digits in $N$.

(1)

(d)

Find the value of $p$ for which $N=\mathrm{e}^{p}$, giving your answer correct to one decimal place.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\log_{10}7^{200}=200\log_{10}7$	(M1)
$169.020$	A1	$169.019608\ldots$
(b)
$N=10^{169.0196\ldots}=10^{0.0196\ldots}\times10^{169}$	(M1)
$a=1.05$	A1	$1.04620\ldots$ Using 169.020 gives 1.047, which also earns A1.
$k=169$	A1	Answer $1.05\times10^{169}$.
(c)
$170$	A1	FT from (b): their k + 1.
(d)
$p=\ln N=200\ln7$	(M1)	Or $p=\log_{10}N\times\ln10$.
$p=389.2$	A1	$389.182\ldots$

Answers: (a) 169.020 · (b) $1.05\times10^{169}$ · (c) 170 · (d) $p=389.2$

+ Add to test✓ In your test78AASL1-q00521.5 Exponents and introduction to logarithms Paper 2Medium13 marks

The magnitude, $M$, of an earthquake on the Richter scale is given by $M=\log_{10}A$, where $A$ is the maximum amplitude, in micrometres, recorded by a seismometer at a standard distance from the centre of the earthquake.

(a)

Find the magnitude of an earthquake for which $A=2.5\times10^{4}$.

(2)

(b)

An earthquake has magnitude 6.3. Find the value of $A$. Give your answer in the form $a\times10^{k}$, where $1\le a\lt 10$ and $k\in\mathbb{Z}$.

(2)

(c)

Two earthquakes have amplitudes $A_{1}$ and $A_{2}$ and magnitudes $M_{1}$ and $M_{2}$. Show that if $M_{2}=M_{1}+1$ then $A_{2}=10A_{1}$.

(2)

(d)

Earthquake X has magnitude 7.1. The amplitude of earthquake Y is 40 times the amplitude of earthquake X. Find the magnitude of earthquake Y.

(3)

(e)

The energy, $E$ joules, released by an earthquake of magnitude $M$ satisfies $\log_{10}E=4.8+1.5M$. Find the energy released by an earthquake of magnitude 6.3.

(2)

(f)

Find how many times more energy is released by earthquake Y than by earthquake X.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$\log_{10}\left(2.5\times10^{4}\right)$	(M1)
$M=4.40$	A1	$4.39794\ldots$
(b)
$A=10^{6.3}$	(M1)
$2.00\times10^{6}$	A1	$1\,995\,262.3\ldots$ Accept $1.995\times10^{6}$. Award A0 for 2E6 or for an answer not in the required form.
(c)
$M_{2}-M_{1}=\log_{10}A_{2}-\log_{10}A_{1}=\log_{10}\dfrac{A_{2}}{A_{1}}$	M1	Or $A_{2}=10^{M_{1}+1}=10^{M_{1}}\times10$.
$\log_{10}\dfrac{A_{2}}{A_{1}}=1$, so $\dfrac{A_{2}}{A_{1}}=10^{1}$	A1
$A_{2}=10A_{1}$	AG
(d)
$A_{Y}=40A_{X}$, so $M_{Y}=\log_{10}\left(40A_{X}\right)$	(M1)
$\log_{10}40+\log_{10}A_{X}=\log_{10}40+7.1$	(M1)	Or $A_{X}=10^{7.1}$ and $\log_{10}\left(40\times10^{7.1}\right)$.
$M_{Y}=8.70$	A1	$8.70205\ldots$
(e)
$\log_{10}E=4.8+1.5(6.3)=14.25$, so $E=10^{14.25}$	(M1)
$1.78\times10^{14}$ J	A1	$1.778\ldots\times10^{14}$. Accept 177 800 000 000 000.
(f)
$\dfrac{E_{Y}}{E_{X}}=10^{1.5\left(M_{Y}-M_{X}\right)}=10^{1.5\times1.60206\ldots}$	(M1)	Or $40^{1.5}$. Accept $10^{4.8+1.5(8.70206)}\div10^{4.8+1.5(7.1)}$.
253 times	A1	$252.98\ldots$ Accept 251 to 254 (251 arises from using 8.70); FT from (d).

Answers: (a) 4.40 · (b) $2.00\times10^{6}$ · (d) 8.70 · (e) $1.78\times10^{14}$ J · (f) 253 times

+ Add to test✓ In your test79AASL1-q00711.7 Rational exponents and laws of logarithms Paper 2Medium7 marks

The mass, $m$ grams, of a radioactive isotope remaining after $t$ years is given by $m=120\left(\tfrac{1}{2}\right)^{\frac{t}{28.8}}$, $t\ge0$.

(a)

Find the mass remaining after 50 years.

(2)

(b)

Find the time taken for the mass to fall to 20 grams.

(3)

(c)

The model can also be written as $m=120\mathrm{e}^{-kt}$. Find the value of $k$.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$120\left(\tfrac{1}{2}\right)^{\frac{50}{28.8}}$	(M1)
$36.0$ g	A1	$36.0213\ldots$
(b)
$\left(\tfrac{1}{2}\right)^{\frac{t}{28.8}}=\tfrac{1}{6}$	(M1)
$\dfrac{t}{28.8}=\dfrac{\ln6}{\ln2}$	(M1)	Or graphical / solver method on the GDC.
$74.4$ years	A1	$74.4469\ldots$
(c)
$\mathrm{e}^{-28.8k}=\tfrac{1}{2}$, or $\mathrm{e}^{-k}=\left(\tfrac{1}{2}\right)^{\frac{1}{28.8}}$	(M1)
$k=\dfrac{\ln2}{28.8}=0.0241$	A1	$0.0240676\ldots$

Answers: (a) 36.0 g · (b) 74.4 years · (c) $k=0.0241$

+ Add to test✓ In your test80AASL1-q00771.8 Infinite geometric series Paper 2Medium8 marks

A ball is dropped from a height of 3.2 m onto horizontal ground. After each bounce it rises to 65% of the height from which it last fell.

(a)

Find the maximum height reached by the ball after its fourth bounce.

(2)

(b)

Find the number of the first bounce after which the maximum height of the ball is less than 5 cm.

(3)

(c)

Assuming the ball continues to bounce in this way indefinitely, find the total vertical distance that it travels.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$3.2\times0.65^{4}$	(M1)
$0.571$ m	A1	$0.57122$ m
(b)
$3.2\times0.65^{n}\lt 0.05$	(M1)	Must convert 5 cm to 0.05 m (or 3.2 m to 320 cm).
$n\gt 9.65$, or heights 0.0663 m after 9 bounces and 0.0431 m after 10 bounces	(A1)
the 10th bounce	A1
(c)
recognizing that each rise is also fallen: $3.2+2\left(3.2\times0.65+3.2\times0.65^{2}+\cdots\right)$	(M1)
$\dfrac{3.2\times0.65}{1-0.65}\;(=5.94286\ldots)$	(M1)
$15.1$ m	A1	$15.0857\ldots$

Answers: (a) 0.571 m · (b) the 10th bounce · (c) 15.1 m

+ Add to test✓ In your test81AASL1-q00781.8 Infinite geometric series Paper 2Medium7 marks

Consider the series $\displaystyle\sum_{k=1}^{\infty}200(0.4)^{k-1}$.

(a)

Find the sum to infinity, $S_{\infty}$.

(2)

(b)

Find the sum of the first six terms, $S_{6}$.

(2)

(c)

Find the least value of $n$ for which $S_{\infty}-S_{n}\lt 0.1$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{200}{1-0.4}$	(M1)
$333$	A1	$333.333\ldots$ Accept $\tfrac{1000}{3}$.
(b)
$\dfrac{200\left(1-0.4^{6}\right)}{1-0.4}$	(M1)
$331.968$	A1	Exact value. Accept 332.
(c)
$\dfrac{1000}{3}-\dfrac{200\left(1-0.4^{n}\right)}{0.6}\lt 0.1$, i.e. $\dfrac{1000}{3}(0.4)^{n}\lt 0.1$	(M1)
$n\gt 8.85$, or $S_{\infty}-S_{8}=0.218$ and $S_{\infty}-S_{9}=0.0874$	(A1)
$n=9$	A1

Answers: (a) $\tfrac{1000}{3}\approx333$ · (b) 331.968 · (c) $n=9$

+ Add to test✓ In your test82AASL1-q00871.9 The binomial theorem Paper 2Medium5 marks

This question uses the binomial coefficient $\dbinom{8}{r}$, where $r\in\mathbb{Z},\ 0\le r\le8$.

(a)

Find the values of $r$ for which $\dbinom{8}{r}=56$.

(2)

(b)

Find the coefficient of $x^{5}$ in the expansion of $(3+2x)^{8}$.

(3)

Show mark scheme

Working	Mark	Notes
(a)
table of values of $\dbinom{8}{r}$ for $r=0,\ldots,8$ (or equivalent)	(M1)
$r=3$ or $r=5$	A1	Both values required.
(b)
$\dbinom{8}{5}(3)^{3}(2x)^{5}$	(M1)	Award for the correct term, with or without $x^{5}$.
$56\times27\times32$	(A1)
$48\,384$	A1

Answers: (a) $r=3$ or $r=5$ · (b) 48 384

+ Add to test✓ In your test83AASL1-q00881.9 The binomial theorem Paper 2Medium6 marks

Consider the expansion of $\left(1+\dfrac{x}{2}\right)^{12}$.

(a)

Write down an expression, in terms of $r$, for the coefficient of $x^{r}$.

(1)

(b)

Find the coefficient of $x^{3}$.

(2)

(c)

Using a table of values, find the value of $r$ for which the coefficient of $x^{r}$ is greatest, and state this greatest coefficient.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\dbinom{12}{r}\left(\dfrac{1}{2}\right)^{r}$	A1
(b)
$\dbinom{12}{3}\left(\dfrac{1}{2}\right)^{3}=\dfrac{220}{8}$	(M1)
$27.5$	A1
(c)
table of $\dbinom{12}{r}\left(\tfrac{1}{2}\right)^{r}$, e.g. $r=3: 27.5,\ r=4: 30.9375,\ r=5: 24.75$	(M1)
$r=4$	A1
$30.9375$	A1	Accept $\tfrac{495}{16}$ or 30.9.

Answers: (b) 27.5 · (c) $r=4$, coefficient 30.9375

+ Add to test✓ In your test84AASL1-q00321.3 Geometric sequences and series Paper 2Hard14 marks

A graduate is offered two jobs.

Offer A: a salary of €42 000 in the first year, increasing by €1500 each year.
Offer B: a salary of €40 000 in the first year, increasing by 4.5% each year.

(a)

Find the salary in the fifth year under Offer A.

(2)

(b)

Find the salary in the fifth year under Offer B.

(2)

(c)

Find the total amount earned in the first 10 years under each offer.

(4)

(d)

Find the first year in which the annual salary under Offer B is greater than the annual salary under Offer A.

(3)

(e)

Find the least number of complete years after which the total amount earned under Offer B is greater than the total amount earned under Offer A.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$42\,000+4(1500)$	(M1)
€48 000	A1
(b)
$40\,000(1.045)^{4}$	(M1)
€47 700.74	A1	Accept €47 700.
(c)
$\dfrac{10}{2}\big(2(42\,000)+9(1500)\big)$	(M1)
Offer A: €487 500	A1
$\dfrac{40\,000\left(1.045^{10}-1\right)}{1.045-1}$	(M1)
Offer B: €491 528.37	A1	Accept €492 000.
(d)
$40\,000(1.045)^{n-1}\gt 42\,000+1500(n-1)$	(M1)
year 5: 47 700.74 < 48 000; year 6: 49 847.28 > 49 500	(A1)
year 6	A1
(e)
$\dfrac{40\,000\left(1.045^{n}-1\right)}{0.045}\gt \dfrac{n}{2}\big(84\,000+1500(n-1)\big)$	(M1)
8 years: 375 200.54 < 378 000; 9 years: 432 084.57 > 432 000	(A1)
9 years	A1

Answers: (a) €48 000 · (b) €47 700.74 · (c) A: €487 500, B: €491 528.37 · (d) year 6 · (e) 9 years

+ Add to test✓ In your test85AASL1-q00331.3 Geometric sequences and series Paper 2Hard13 marks

During an outbreak of a disease, 40 new cases were recorded on day 1. For days 1 to 10, the number of new cases each day was 25% greater than on the previous day. From day 11 onwards, after restrictions were introduced, the number of new cases each day was 20% less than on the previous day.

In this question, give numbers of cases to the nearest whole number.

(a)

Find the number of new cases recorded on day 10.

(2)

(b)

Find the total number of cases recorded in the first 10 days.

(2)

(c)

Find the first day at the end of which the total number of cases recorded exceeded 1000.

(3)

(d)

Find the first day on which fewer than 20 new cases were recorded.

(3)

(e)(i)

The total number of cases recorded from day 11 to day 20 inclusive.

Write down an expression, using sigma notation, for this total.

(1)

(e)(ii)

Hence find the total number of cases recorded in the first 20 days.

(2)

Show mark scheme

Working	Mark	Notes
(a)
$40(1.25)^{9}$	(M1)
$298$	A1	$298.02\ldots$
(b)
$\dfrac{40\left(1.25^{10}-1\right)}{1.25-1}$	(M1)
$1330$	A1	$1330.1\ldots$
(c)
$\dfrac{40\left(1.25^{n}-1\right)}{0.25}\gt 1000$	(M1)
$n\gt 8.88$; or $S_{8}=794$, $S_{9}=1032$	(A1)
day 9	A1
(d)
day $10+m$: $298.02(0.8)^{m}\lt 20$	(M1)	FT from (a). Award M0 for using 0.2.
$m\gt 12.1$; or day 22: 20.5, day 23: 16.4	(A1)
day 23	A1
(e)(i)
$\displaystyle\sum_{m=1}^{10}298.02(0.8)^{m}$	A1	Accept equivalent forms, e.g. $\sum_{m=1}^{10}40(1.25)^{9}(0.8)^{m}$ or $\sum_{m=0}^{9}238.42(0.8)^{m}$.
(e)(ii)
$\dfrac{238.42\left(1-0.8^{10}\right)}{1-0.8}\;(=1064.09)$ added to their (b)	(M1)
$2394$	A1	$2394.2\ldots$ Accept 2390 or 2395 from rounded values.

Answers: (a) 298 · (b) 1330 · (c) day 9 · (d) day 23 · (e)(ii) 2394

+ Add to test✓ In your test86AASL1-q00421.4 Financial applications Paper 2Hard6 marks

Bank X offers an interest rate of 4.1% per year, compounded annually. Bank Y offers a nominal annual interest rate of 4.0%, compounded monthly.

(a)

Determine which bank gives the greater value after 5 years for an investment of €10 000. Justify your answer by finding both values to the nearest cent.

(3)

(b)

Bank Y decides to change its nominal annual rate, still compounded monthly, so that its investments grow at exactly the same rate as those at Bank X. Find the new nominal annual rate. Give your answer correct to three significant figures.

(3)

Show mark scheme

Working	Mark	Notes
(a)
Bank X: $10\,000(1.041)^{5}=$€12 225.13	A1
Bank Y: $10\,000\left(1+\dfrac{4}{1200}\right)^{60}=$€12 209.97	A1
Bank X gives the greater value	R1	Must follow from two values.
(b)
$\left(1+\dfrac{r}{1200}\right)^{12}=1.041$	(M1)	Or TVM with any equal starting value.
$r=4.0249\ldots$	(A1)
$4.02\%$	A1

Answers: (a) Bank X (€12 225.13 versus €12 209.97) · (b) 4.02%

+ Add to test✓ In your test87AASL1-q00431.4 Financial applications Paper 2Hard8 marks

Throughout this question, the annual rate of inflation is 2.4%. The real value of an amount after $n$ years, in terms of today’s money, is found by dividing its value after $n$ years by $1.024^{n}$.

(a)

Omar keeps €1500 in cash for 10 years. Find the real value of this cash after 10 years, in terms of today’s money.

(2)

(b)

Find the number of complete years after which the real value of Omar’s €1500 is first less than €1200.

(3)

(c)

Omar instead invests €20 000 for 10 years at a nominal annual rate of $r\%$, compounded monthly. Find the least value of $r$, correct to two decimal places, for which the real value of the investment after 10 years is at least €22 000.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$\dfrac{1500}{1.024^{10}}$	(M1)
€1183.29	A1
(b)
$\dfrac{1500}{1.024^{n}}\lt 1200$	(M1)
$n\gt 9.41$; or after 9 years €1211.69, after 10 years €1183.29	(A1)
10 years	A1
(c)
$\dfrac{20\,000\left(1+\frac{r}{1200}\right)^{120}}{1.024^{10}}\ge22\,000$	(M1)	Accept use of FV $=22\,000\times1.024^{10}=27\,888.31$ in TVM.
$r=3.3293\ldots$	(A1)
$r=3.33$	A1	Award A0 for 3.32 (gives a real value of €21 979.46, which is too small). The approximate method of subtracting the inflation rate from $r$ (not the method given in the stem) gives $3.35$: award (M1)(A0)A0.

Answers: (a) €1183.29 · (b) 10 years · (c) $r=3.33$

+ Add to test✓ In your test88AASL1-q00441.4 Financial applications Paper 2Hard13 marks

At the start of each year, Priya deposits $2000 into a savings account that pays interest at 3% per year, compounded annually. She makes no withdrawals. Let $V_{n}$ be the value of the account at the end of year $n$, immediately after interest has been added.

(a)

Show that $V_{2}=4181.80$.

(2)

(b)

Show that $V_{n}=\dfrac{206\,000}{3}\left(1.03^{n}-1\right)$.

(3)

(c)

Find the value of $V_{10}$.

(2)

(d)

Find the least value of $n$ for which $V_{n}\gt 50\,000$.

(3)

(e)

Instead, Priya could deposit a single amount of $20 000 at the start of year 1 in the same account. Find how much greater the value of the account would be at the end of year 10 under this option.

(3)

Show mark scheme

Working	Mark	Notes
(a)
$V_{1}=2000\times1.03=2060$, then $V_{2}=(2060+2000)\times1.03$	M1
$4060\times1.03$	A1
$=4181.80$	AG
(b)
$V_{n}=2000(1.03)^{n}+2000(1.03)^{n-1}+\cdots+2000(1.03)$	M1	Recognizing a geometric series.
$u_{1}=2000(1.03),\ r=1.03$, so $V_{n}=\dfrac{2060\left(1.03^{n}-1\right)}{1.03-1}$	A1
$\dfrac{2060}{0.03}=\dfrac{206\,000}{3}$	A1
$V_{n}=\dfrac{206\,000}{3}\left(1.03^{n}-1\right)$	AG
(c)
$\dfrac{206\,000}{3}\left(1.03^{10}-1\right)$	(M1)
$23 615.59	A1
(d)
$\dfrac{206\,000}{3}\left(1.03^{n}-1\right)\gt 50\,000$	(M1)
$n\gt 18.5$; or $V_{18}=48\,233.74$, $V_{19}=51\,740.75$	(A1)
$n=19$	A1
(e)
$20\,000(1.03)^{10}$	(M1)
$26 878.33	(A1)
$26\,878.33-23\,615.59=$$3262.74	A1	FT from (c).

Answers: (c) $23 615.59 · (d) $n=19$ · (e) $3262.74

Working	Mark	Notes
(a)
\(3.6\times1.2=4.32\) and \(10^{5}\times10^{-2}=10^{3}\)	(M1)	Award for either product of the decimal parts or correct addition of indices.
\(4.32\times10^{3}\)	A1
(b)
\(3.6\div1.2=3\) and \(10^{5}\div10^{-2}=10^{7}\)	(M1)
\(3\times10^{7}\)	A1	Accept \(3.0\times10^{7}\). Award A0 for \(30\times10^{6}\) or calculator notation such as 3E7.

Working	Mark	Notes
(a)
\(4.07\times10^{-4}\)	A1
(b)
\(5.23\times10^{5}\)	A1
(c)
writing all four in a common form, e.g. \(0.031,\ 0.0298,\ 0.0312,\ 0.0299\)	(M1)	Or all in the form \(a\times10^{-2}\).
\(0.0298,\ \ 2.99\times10^{-2},\ \ 3.1\times10^{-2},\ \ 31.2\times10^{-3}\)	A1

Working	Mark	Notes
(a)
\(640\,000+38\,000\) or \(6.4\times10^{5}+0.38\times10^{5}\)	(M1)	Award for writing both numbers with the same power of 10 or as ordinary numbers.
\(6.78\times10^{5}\)	A1
(b)
\(10b=3.8\times10^{5}\), so \((6.4-3.8)\times10^{5}\)	(M1)
\(2.6\times10^{5}\)	A1
(c)
\(a^{2}=40.96\times10^{10}\)	(M1)	Or \(2\times6.4^{2}=81.92\).
\(8.192\times10^{11}\)	A1	Award A0 for \(81.92\times10^{10}\) (not in the required form).

Working	Mark	Notes
(a)
\(u_{1}+3d=17\) and \(u_{1}+8d=42\), or \(5d=42-17\)	(M1)
\(d=5\)	A1
(b)
\(u_{1}=17-3(5)=2\)	A1	FT from their \(d\).
(c)
\(S_{20}=\dfrac{20}{2}\big(2(2)+19(5)\big)\)	(M1)	Correct substitution of their values into either sum formula.
\(990\)	A1

Working	Mark	Notes
(a)
\(d=-6\)	A1
(b)
\(u_{n}=23+(n-1)(-6)\)	(M1)
\(u_{n}=29-6n\)	A1
(c)
\(29-6n=-121\)	(M1)	FT from (b).
\(n=25\)	A1
(d)
\(29-6n=-100\Rightarrow n=21.5\)	M1
\(n\) is not an integer, so \(-100\) is not a term	R1	Do not award R1 without a valid method.

Working	Mark	Notes
(a)
\(4000+6d=4780\)	(M1)
$130	A1
(b)
\(\dfrac{130}{4000}\times100\)	(M1)	FT from (a).
\(3.25\%\)	A1
(c)
\(4000+130n\gt 6000\Rightarrow n\gt 15.38\ldots\)	(M1)
16 years	A1

Working	Mark	Notes
(a)
\(\dfrac{48\,000-37\,200}{3}\)	(M1)
$3600	A1
(b)
\(48\,000-10(3600)=12\,000\), so $12 000	A1	FT from (a).
(c)
10% of the price is $4800	(A1)
\(48\,000-3600n\lt 4800\Rightarrow n\gt 12\)	(M1)	After 12 years the value is exactly $4800.
13 years	A1	Award A0 for 12.

Working	Mark	Notes
(a)
\(r=\dfrac{1}{3}\)	A1
(b)
\(2187\left(\dfrac{1}{3}\right)^{7}\)	(M1)
\(u_{8}=1\)	A1
(c)
\(u_{n}=3^{7}\left(\dfrac{1}{3}\right)^{n-1}\)	M1
\(3^{7}\times3^{-(n-1)}=3^{7-n+1}\)	A1
\(u_{n}=3^{8-n}\)	AG
(d)
\(3^{8-n}=3^{-4}\)	(M1)
\(n=12\)	A1

Working	Mark	Notes
(a)
\(5000\times1.1^{2}\)	(M1)
€6050	A1
(b)
\(20\,000\times0.75^{2}\)	(M1)	Award M0 for \(20\,000\times0.5\).
€11 250	A1
(c)
\(\dfrac{20\,000-11\,250}{20\,000}\times100=43.75\%\)	A1	FT from (b). Not 50%.

Working	Mark	Notes
(a)
\((2x^{3}y^{-2})^{4}=16x^{12}y^{-8}\)	(M1)
\(2x^{7}\)	A1
\(y^{-5}\), giving \(2x^{7}y^{-5}\)	A1
(b)
\(2\times2^{7}\times2^{-5}=8\)	A1	FT from their simplified expression.

Working	Mark	Notes
(a)
\(k^{2}-(k+5)=(5k+3)-k^{2}\), or \(2k^{2}=(k+5)+(5k+3)\)	M1
\(2k^{2}=6k+8\)	A1
\(k^{2}-3k-4=0\)	AG
(b)
\((k-4)(k+1)=0\), so \(k=4\) or \(k=-1\)	A1
\(k=-1\) gives terms \(4,\ 1,\ -2\) (\(d=-3\lt 0\)); \(k=4\) gives \(9,\ 16,\ 23\) (\(d=7\gt 0\))	R1	Must consider the sign of \(d\) for each value.
\(k=4\)	A1
(c)
\(d=7\) and \(u_{1}=9-7=2\)	A1
\(S_{20}=\dfrac{20}{2}\big(2(2)+19(7)\big)\)	(M1)	FT from their \(u_{1}\) and \(d\).
\(1370\)	A1

Working	Mark	Notes
(a)
\(0.36+0.0036+0.000036\)	A1
\(r=0.01\)	A1
(b)
\(\dfrac{0.36}{1-0.01}=\dfrac{36}{99}\)	(M1)
\(\dfrac{4}{11}\)	A1
(c)
\(0.1+(0.027+0.00027+\cdots)\)	M1
\(0.1+\dfrac{0.027}{0.99}=\dfrac{1}{10}+\dfrac{27}{990}\)	M1	Or \(\tfrac{1}{10}\left(1+\tfrac{27}{99}\right)\).
\(\dfrac{7}{55}\)	A1

Working	Mark	Notes
(a)
\(-3\)	A1
(b)
\(-2\)	A1
(c)
\(2x-1=3\), or \(2x-1=\log_{10}1000\)	(M1)
\(x=2\)	A1

Working	Mark	Notes
(a)
\(\log_{10}0.01=-2\)	A1	Accept \(\lg 0.01=-2\) or \(\log 0.01=-2\).
(b)
\(y=\mathrm{e}^{3}\)	A1
(c)
\(a^{4}=81\)	(M1)
\(a=3\)	A1	Award A0 if \(a=-3\) is also given.
(d)
\(10^{x}\gt 0\) for all real \(x\) (equivalently, \(\log_{10}b\) is defined only for \(b\gt 0\))	R1	Accept any correct reason referring to the positivity of the power or to the domain of the logarithm.

Working	Mark	Notes
(a)
\(8^{-2}=2^{-6}\) and \(4^{3}=2^{6}\)	(M1)	Award for both conversions to base 2.
\(2^{-6+6-(-5)}=2^{5}\)	A1
(b)
\(5^{n-2}=5^{-3}\)	(M1)
\(n=-1\)	A1

Working	Mark	Notes
(a)
\(64-49=15\) and \(8+7=15\)	A1	Both sides must be evaluated.
(b)
\(n^{2}+2n+1-n^{2}\)	M1	Must start from the LHS.
\(2n+1\)	A1	Final line must show the simplification, not just the answer.
\((n+1)^{2}-n^{2}\equiv2n+1\)	AG
(c)
\(2n+1=41\), so \(n=20\)	(M1)
\(41=21^{2}-20^{2}\)	A1

Working	Mark	Notes
(a)
LHS \(=9\), RHS \(=5\), and \(9\neq5\)	A1
(b)
\(x^{2}+4x+4=x^{2}+4\), so \(4x=0\)	(M1)
\(x=0\) only	A1
(c)
\(\left(x^{2}+4x+4\right)-\left(x^{2}-4x+4\right)\)	M1	Brackets (or correct signs) required.
\(=x^{2}+4x+4-x^{2}+4x-4\)	A1
\(\equiv8x\)	AG
(d)
LHS \(=49-9=40\); RHS \(=8\times5=40\)	A1

Working	Mark	Notes
(a)
\(8\)	A1
(b)
\(\left(\dfrac{27}{8}\right)^{\frac{2}{3}}\) or \(\left(\dfrac{2}{3}\right)^{-2}\)	(M1)
\(\dfrac{9}{4}\)	A1
(c)
\(x=64^{\frac{2}{3}}\)	(M1)
\(x=16\)	A1

Working	Mark	Notes
(a)
\(2\log_{6}3=\log_{6}9\)	M1
\(\log_{6}9+\log_{6}4=\log_{6}36\)	M1
\(2\)	A1
(b)
\(\ln8-\ln4+\ln5=\ln\dfrac{8\times5}{4}\)	(M1)	Award for at least two correct uses of the laws.
\(\ln10\)	A1
(c)
\(0+1=1\)	A1

Working	Mark	Notes
(a)
\(r=-\tfrac{1}{2}\)	A1
(b)
\(\dfrac{24}{1-\left(-\frac{1}{2}\right)}\)	(M1)
\(16\)	A1
(c)
\(\dfrac{u_{1}}{1-\frac{1}{3}}=45\)	(M1)
\(u_{1}=30\)	A1

Working	Mark	Notes
(a)
\(p=16\times10^{8}\), so \(\sqrt{p}=\sqrt{16}\times\sqrt{10^{8}}\)	(M1)
\(4\times10^{4}\)	A1
(b)
\(q^{2}=16\times10^{-6}=1.6\times10^{-5}\)	(M1)
\(2.56\times10^{4}\)	A1
(c)
\(q^{3}=64\times10^{-9}=6.4\times10^{-8}\)	(M1)
\(\dfrac{1.6\times10^{9}}{6.4\times10^{-8}}=0.25\times10^{17}=2.5\times10^{16}\)	(A1)
\(n=16\)	A1

Working	Mark	Notes
(a)
\(3+7+11\)	A1	Accept 3, 7, 11.
(b)
recognizing an arithmetic series with \(u_{1}=3,\ d=4\) and substituting: \(\dfrac{n}{2}\big(2(3)+4(n-1)\big)\)	M1
\(\dfrac{n}{2}(4n+2)\) or \(n(2n+1)\)	A1
\(=2n^{2}+n\)	AG
(c)
\(2n^{2}+n-210=0\)	M1
\((2n+21)(n-10)=0\)	A1	Or correct substitution into the quadratic formula.
\(n=10\)	A1	Award the final A1 only if \(n=-\tfrac{21}{2}\) is rejected or not given.

Working	Mark	Notes
(a)
\(36\)	A1
(b)
first term \(17\), last term \(122\)	A1	Or \(u_{1}=17,\ d=3\).
\(\dfrac{36}{2}(17+122)\)	(M1)	FT from their number of terms.
\(2502\)	A1
(c)
\(5r+3\)	A1
\(\displaystyle\sum_{r=1}^{40}(5r+3)\)	A1	Upper limit 40 required.

Working	Mark	Notes
(a)
\(u_{1}r=12\) and \(u_{1}r^{4}=-324\), so \(r^{3}=-27\)	(M1)
\(r=-3\)	A1
(b)
\(u_{1}=-4\)	A1	FT from their \(r\).
(c)
\(S_{6}=\dfrac{-4\big((-3)^{6}-1\big)}{-3-1}\)	(M1)
\(728\)	A1
(d)
\(\|r\|=3\gt 1\), so the series diverges	R1	Must refer to the value of \(\|r\|\) or \(r\).

Working	Mark	Notes
(a)
\(\dfrac{k-2}{k-4}=\dfrac{2k-7}{k-2}\), or \((k-2)^{2}=(k-4)(2k-7)\)	M1
\(k^{2}-4k+4=2k^{2}-15k+28\)	A1
\(k^{2}-11k+24=0\)	AG
(b)
\((k-3)(k-8)=0\), so \(k=3\) or \(k=8\)	A1
\(k=3\) gives \(-1,\ 1,\ -1\), which are not all positive	R1
\(k=8\), terms \(4,\ 6,\ 9\), \(r=\dfrac{3}{2}\)	A1	Both \(k\) and \(r\) required.
(c)
\(4\left(\dfrac{3}{2}\right)^{4}\)	(M1)	FT from (b).
\(\dfrac{81}{4}\)	A1	Accept 20.25.

Working	Mark	Notes
(a)
\(6+12+24\)	A1
(b)
\(\dfrac{6\left(2^{8}-1\right)}{2-1}\)	(M1)	Recognizes \(u_{1}=6,\ r=2,\ n=8\).
\(1530\)	A1
(c)
\(1530-(6+12+24)\)	(M1)	Must use (b).
\(1488\)	A1

Working	Mark	Notes
(a)
\(10^{3}=1000\lt 4500\lt 10\,000=10^{4}\), so \(3\lt\log_{10}4500\lt 4\)	R1
3 and 4	A1
(b)
\(7\)	A1
(c)
\(2x=\ln5\)	(M1)
\(x=\tfrac{1}{2}\ln5\)	A1
(d)
\(\mathrm{e}\lt 2.8\lt 5\) and \(\mathrm{e}^{2}\gt 2.7^{2}=7.29\gt 5\), so \(\mathrm{e}^{1}\lt 5\lt\mathrm{e}^{2}\)	M1	Must see a numerical comparison of 5 with both e and e².
hence \(1\lt\ln5\lt 2\), so \(\tfrac{1}{2}\lt\tfrac{1}{2}\ln5\lt 1\)	R1

Working	Mark	Notes
(a)
\(\dfrac{(n+2)-n}{n(n+2)}\)	M1	Must start from the LHS and use a common denominator.
numerator simplified to \(2\)	A1
\(\equiv\dfrac{2}{n(n+2)}\)	AG
(b)
\(n(n+2)=63\), so \(n=7\)	(M1)
\(\dfrac{1}{7}-\dfrac{1}{9}\)	A1

Working	Mark	Notes
(a)
expanding at least two of the brackets correctly	M1
\(n^{2}-2n+1+n^{2}+n^{2}+2n+1\)	A1
\(\equiv3n^{2}+2\)	AG
(b)
\(3n^{2}\) is a multiple of 3 for every integer \(n\)	R1
so the sum is always 2 more than a multiple of 3, hence it is never a multiple of 3	R1
(c)
\(3n^{2}+2=110\), so \(n^{2}=36\)	(M1)
5, 6, 7 and \(-7,\,-6,\,-5\)	A1	Both sets required for A1.

Working	Mark	Notes
(a)
RHS \(=2\left(x^{2}-6x+9\right)+5\)	M1	Working from either side is acceptable; must show the expansion or the completion of the square.
\(=2x^{2}-12x+18+5\)	A1
\(=2x^{2}-12x+23\)	AG
(b)
\((x-3)^{2}\ge0\) for all real \(x\)	R1
so \(2(x-3)^{2}+5\ge5\gt 0\)	R1
(c)
denominator has least value 5	(M1)
greatest value 2, when \(x=3\)	A1	Both required.

Working	Mark	Notes
(a)
\(4n^{2}+4n+1\)	M1
\(=4\left(n^{2}+n\right)+1\)	A1
\(\equiv4n(n+1)+1\)	AG
(b)
\(n\) and \(n+1\) are consecutive integers, so one of them is even	R1
(c)
\(n(n+1)=2k\) for some \(k\in\mathbb{Z}\), so \(4n(n+1)=8k\)	R1
every odd integer can be written as \(2n+1\), so its square is \(8k+1\), one more than a multiple of 8	R1
(d)
\(1\)	A1

Working	Mark	Notes
(a)
\(\log_{3}a^{2}+\log_{3}b\)	(M1)
\(2p+q\)	A1
(b)
\(\log_{3}9+\log_{3}a-\tfrac{1}{2}\log_{3}b\)	(M1)	Award for two correct uses of the laws of logarithms.
\(2+p-\dfrac{q}{2}\)	A1
(c)
\(\dfrac{\log_{3}b}{\log_{3}9}\)	(M1)
\(\dfrac{q}{2}\)	A1