This note covers arithmetic sequences, where a fixed amount is added each time, and geometric sequences, where each term is multiplied by a fixed ratio. It shows how to find a term, a sum, and how to model situations such as savings plans, seating plans and steady percentage growth.
Key ideas
A sequence is an ordered list of numbers. The terms are written \(u_1, u_2, u_3,\dots\) and \(u_n\) is the \(n\)th term. A series is the sum of the terms, and \(S_n\) is the sum of the first \(n\) terms.
Arithmetic sequences. The difference between consecutive terms is a constant \(d\), called the common difference. The sequence \(7, 11, 15, 19,\dots\) has \(d=4\). It grows (or falls, if \(d\) is negative) by the same amount each step, so it is a linear model.
\[ u_n = u_1 + (n-1)d \qquad S_n = \frac{n}{2}\bigl(2u_1+(n-1)d\bigr) = \frac{n}{2}(u_1+u_n) \]Geometric sequences. The ratio of consecutive terms is a constant \(r\), called the common ratio. The sequence \(3, 6, 12, 24,\dots\) has \(r=2\). A sequence with \(r=1.05\) grows by 5% each step, and one with \(r=0.9\) falls by 10% each step. It is an exponential model.
\[ u_n = u_1 r^{\,n-1} \qquad S_n = \frac{u_1(r^{n}-1)}{r-1} = \frac{u_1(1-r^{n})}{1-r}\quad (r\ne1) \]Identifying the type. Subtract consecutive terms. If the differences are equal, the sequence is arithmetic. Divide consecutive terms. If the ratios are equal, it is geometric. Check at least three terms before deciding.
Finding \(u_1\) and \(d\) (or \(r\)) from two terms. Write each given term using the formula. This gives two equations. For an arithmetic sequence, subtract them to find \(d\). For a geometric sequence, divide them to find \(r^{\,\text{something}}\) and then take a root.
Finding \(n\). To find when a sequence reaches a value, write an equation or inequality in \(n\). For an arithmetic sequence you can solve it by algebra. For a geometric sequence, either use logarithms or use the table of values on your GDC and look for where the term first passes the target. You can also use the GDC to find the sum of a sequence, for instance by building a table of \(u_n\) and using the sum feature, which is useful to check your formula work.
Context. The first term is the value at the start. Take care with what \(n=1\) means. If a savings plan starts with a deposit in January, then January is \(n=1\) and the next January is \(n=13\), not \(n=12\).
Round only at the end. Sequence questions about people or objects need whole-number answers for \(n\), and an inequality such as \(3\times2^{n-1}\gt1000\) means you give the first whole \(n\) that works.
Linking to other models. The terms of an arithmetic sequence lie on a straight line when plotted against \(n\), with gradient \(d\). The terms of a geometric sequence lie on an exponential curve. This is why simple interest produces an arithmetic sequence of balances and compound interest produces a geometric one.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(d = u_{n+1}-u_n\) | Common difference of an arithmetic sequence. |
| \(u_n = u_1+(n-1)d\) | The \(n\)th term of an arithmetic sequence. |
| \(S_n = \dfrac{n}{2}\bigl(2u_1+(n-1)d\bigr)\) | Sum of the first \(n\) terms of an arithmetic sequence. Use \(\dfrac{n}{2}(u_1+u_n)\) if you know the last term. |
| \(r = \dfrac{u_{n+1}}{u_n}\) | Common ratio of a geometric sequence. |
| \(u_n = u_1r^{\,n-1}\) | The \(n\)th term of a geometric sequence. |
| \(S_n = \dfrac{u_1(r^{n}-1)}{r-1}\) | Sum of the first \(n\) terms of a geometric sequence, \(r\ne1\). |
| \(r = 1+\dfrac{p}{100}\) | Ratio for a percentage increase of \(p\%\) each step. Use \(1-\dfrac{p}{100}\) for a decrease. |
Worked examples
Example 1 The first row of a stadium stand has 18 seats. Each row behind has 3 more seats than the row in front. There are 24 rows. Find the number of seats in the last row and the total number of seats.
This is arithmetic with \(u_1=18\), \(d=3\), \(n=24\).
\(u_{24} = 18 + (24-1)\times3 = 18+69 = 87\).
\(S_{24} = \dfrac{24}{2}(18+87) = 12\times105 = 1260\).
Answer: 87 seats in the last row; 1260 seats in total.
Example 2 A saver puts £5 into a jar in the first month and increases the monthly amount by 8% each month. Find the amount saved in month 10, and the total saved in the first 12 months, giving both to the nearest penny.
An 8% increase each month means \(r=1.08\). The first term is \(u_1=5\).
Month 10: \(u_{10} = 5\times1.08^{9} = 9.995\ldots\), which is £10.00 to the nearest penny.
Total after 12 months: \(S_{12} = \dfrac{5(1.08^{12}-1)}{1.08-1} = \dfrac{5\times0.51817\ldots}{0.08} = 94.885\ldots\)
Answer: £10.00 in month 10; £94.89 saved in total.
Example 3 An arithmetic sequence has \(u_3=19\) and \(u_8=44\). Find \(u_1\) and \(d\), then find the sum of the first 20 terms.
Write the two terms: \(u_1+2d = 19\) and \(u_1+7d = 44\).
Subtract the first from the second: \(5d = 25\), so \(d = 5\).
Substitute: \(u_1 + 10 = 19\), so \(u_1 = 9\).
\(S_{20} = \dfrac{20}{2}\bigl(2\times9+19\times5\bigr) = 10\times113 = 1130\).
Answer: \(u_1=9\), \(d=5\), \(S_{20}=1130\).
Common errors
- Common error: using \(u_1+nd\) instead of \(u_1+(n-1)d\). The first term has no differences added.
- Common error: using \(r=0.08\) for an 8% increase. The ratio is \(1.08\), and for an 8% decrease it is \(0.92\).
- Common error: writing \(u_1r^{n}\) as the \(n\)th term. The power is \(n-1\).
- Common error: calling a sequence arithmetic after checking only the first two terms. Always test a third.
- Common error: giving a non-whole value of \(n\) when the question asks "how many terms" or "which month".
Exam tip: state the type of sequence and the values of \(u_1\), \(d\) or \(r\), and \(n\) before substituting into a formula, so that each step is easy to follow.
Check yourself
- Find the sum of the first 20 terms of \(7, 11, 15, \dots\)
Answer
\(u_1=7\), \(d=4\), \(n=20\). \(S_{20}=\dfrac{20}{2}(2\times7+19\times4)=10\times90=900\). - A geometric sequence has \(u_1=500\) and \(r=0.9\). Find \(u_6\) to 3 s.f.
Answer
\(u_6 = 500\times0.9^{5} = 295.245\) which is \(295\) to 3 s.f. - Find the first term of \(3, 6, 12, \dots\) that is greater than 1000.
Answer
\(u_n=3\times2^{n-1}\). For \(n=9\) it is \(768\); for \(n=10\) it is \(1536\). So the 10th term, \(1536\), is the first above 1000. - Find the sum of the first 8 terms of \(2, 6, 18, \dots\)
Answer
\(r=3\). \(S_8=\dfrac{2(3^{8}-1)}{3-1}=3^{8}-1=6560\).