Edexcel International A Level Maths · P2 · Pure Mathematics 2
Sequences and series
28 free, high-quality past-paper-style questions on Sequences and series for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 254 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-3-B-01P2 · ASEasy6 marks
The seats in a small theatre are arranged in 25 rows. The front row has 18 seats, and each row after the first has 2 more seats than the row in front of it.
(a)
Find the number of seats in the back row (row 25).
(2)
(b)
Find the total number of seats in the theatre.
(2)
(c)
Find the number of the first row that has more than 60 seats.
(2)
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Answer
Mark
Notes
Part (a)
uses un = a + (n − 1)d with a = 18, d = 2, n = 25
M1
18 + 24 × 2 = 66 seats
A1
Part (b)
uses Sn = ½n(a + l) or ½n[2a + (n − 1)d] with n = 25
M1
½ × 25 × (18 + 66) = 1050 ft on their 66
A1
Part (c)
18 + 2(n − 1) > 60, giving n > 22
M1
row 22 has exactly 60 seats, so row 23
A1
Common error: Answering 22. Row 22 has exactly 60 seats, which is not more than 60. Check the boundary value whenever the inequality is strict.
A toy pendulum is released and swings to and fro. The first swing moves the bob through an arc of length 40 cm. A model assumes that the arc length of each swing after the first is ¾ of the arc length of the swing before it.
(a)
Find the exact arc length of the 6th swing, according to the model.
(2)
(b)
Find the total arc length of the first 10 swings, giving your answer to 3 significant figures.
(2)
(c)
Show that, according to the model, the total arc length can never reach 160 cm.
(2)
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Answer
Mark
Notes
Part (a)
uses arn−1 with a = 40, r = ¾, n = 6
M1
40 × (¾)5 = 1215/128 cm
A1
Part (b)
uses a(1 − rn)/(1 − r) with n = 10
M1
160(1 − 0.7510) = 150.99… = 151 cm
A1
Part (c)
|r| = ¾ < 1, so S∞ = 40 ÷ (1 − ¾) = 160
M1
Sn = 160(1 − 0.75n) and 0.75n > 0 for every n, so every total is less than 160 cm
A1
Exam tip: “Exact” means leave the answer as a fraction (or surd); do not round it. 40 × 243/1024 simplifies to 1215/128.
Exam tip: In a “show that”, the answer is printed, so the working is what earns the marks. Write the sum formula with the values substituted before you simplify.
Figure 1 shows the first three patterns in a sequence of matchstick patterns. Pattern n is a row of n triangles, and each side of each triangle is one matchstick.
(a)
Use Figure 1 to explain why the numbers of matchsticks in the patterns form an arithmetic sequence, and state its first term and common difference.
(2)
(b)
Find the number of matchsticks in pattern 40.
(1)
(c)
Lena has 500 matchsticks. She makes pattern 1, then pattern 2, then pattern 3, and so on, each as a separate pattern, until she does not have enough matchsticks left for the next pattern. Find the number of patterns she makes and the number of matchsticks left over.
(3)
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Answer
Mark
Notes
Part (a)
the patterns use 3, 5, 7 matchsticks: each extra triangle shares a side with the one before, so it needs 2 more matchsticks every time
B1
first term 3, common difference 2
B1
Part (b)
3 + 39 × 2 = 81
B1
Part (c)
total for patterns 1 to n: ½n[6 + 2(n − 1)] = n(n + 2)
M1
21 × 23 = 483 ≤ 500 but 22 × 24 = 528 > 500
M1
21 patterns, with 17 matchsticks left over
A1
Common error: Counting 3 matchsticks for every triangle. Neighbouring triangles share a side, so each new triangle needs only 2 more matchsticks.
An app is released. Figure 1 shows the number of downloads, in thousands, in each of its first four weeks. A model assumes that the numbers of downloads in weeks 1, 2, 3, … form a geometric sequence.
(a)
Use Figure 1 to show that the common ratio of the sequence is 1.5.
(2)
(b)
Find, according to the model, the total number of downloads in the first 10 weeks. Give your answer to the nearest thousand.
(2)
(c)
Find the first week in which, according to the model, there are more than 1 million downloads in that week.
(3)
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Answer
Mark
Notes
Part (a)
reads 8, 12, 18 and 27 (thousand) from the graph
B1
12 ÷ 8 = 18 ÷ 12 = 27 ÷ 18 = 1.5
B1
Part (b)
8(1.510 − 1)/(1.5 − 1)
M1
= 906.64… thousand, so 907 000 downloads
A1
Part (c)
8 × 1.5n−1 > 1000, so 1.5n−1 > 125
M1
n − 1 > log 125 / log 1.5 = 11.9 (3 s.f.)
M1
n = 13: week 13
A1
Exam tip: Part (c) is about the downloads in a single week (one term), not the running total (a sum). Decide which one the question wants before choosing a formula.
Figure 1 shows the first three patterns in a sequence of dot patterns. Each pattern is a rectangle of dots, and pattern n has n rows. The number of dots in pattern n is un.
(a)
Find u4, and find an expression for un in terms of n.
(2)
(b)
Find which pattern contains 360 dots.
(2)
(c)
Show that un+1 − un = 2n + 3, and hence explain why the sequence is increasing.
(3)
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Answer
Mark
Notes
Part (a)
u4 = 4 × 6 = 24
B1
pattern n has n rows of n + 2 dots, so un = n(n + 2)
B1
Part (b)
n2 + 2n − 360 = 0
M1
(n − 18)(n + 20) = 0, so pattern 18
A1
Part (c)
(n + 1)(n + 3) − n(n + 2)
M1
= n2 + 4n + 3 − n2 − 2n = 2n + 3
A1
2n + 3 > 0 for every n ≥ 1, so each term is greater than the one before: the sequence is increasing
B1
Common error: Treating the sequence as arithmetic. The increases are 5, 7, 9, …, which are not constant, so build the rule from rows × dots per row instead.
Figure 1 shows the first eight terms of a sequence u1, u2, u3, …, plotted against n. The sequence is defined by un+1 = pun + q, where p and q are constants.
(a)
Use Figure 1 to write down the values of u1, u2 and u3.
(1)
(b)
Find the value of p and the value of q.
(3)
(c)
Find Σr=135ur.
(2)
(d)
Find the value of u1 for which this recurrence relation would give a constant sequence.
(1)
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Answer
Mark
Notes
Part (a)
u1 = −2, u2 = 7, u3 = −2
B1
Part (b)
7 = −2p + q
M1
−2 = 7p + q
M1
subtracting: 9 = −9p, so p = −1 and q = 5
A1
Part (c)
the sequence has period 2 and each pair sums to −2 + 7 = 5; 35 terms = 17 pairs + u35 = −2
M1
17 × 5 − 2 = 83
A1
Part (d)
u1 = 5 − u1, so u1 = 2.5
B1
Exam tip: A sequence with period 2 can be summed in pairs: count the complete pairs, then add any term left over at the end.
Ravi repays a loan in monthly instalments. His first instalment is $150, and each instalment after the first is $d more than the one before. The total of his first 24 instalments is $6360.
(a)
Show that d = 10.
(3)
(b)
Find the amount of his 24th instalment.
(1)
(c)
The loan is fully repaid when the total of his instalments first reaches at least $12 000. Find the number of instalments needed.
(4)
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Answer
Mark
Notes
Part (a)
uses ½n[2a + (n − 1)d] with a = 150 and n = 24
M1
12(300 + 23d) = 6360
M1
300 + 23d = 530, so d = 10
A1
Part (b)
150 + 23 × 10 = $380
B1
Part (c)
½n[300 + 10(n − 1)] ≥ 12 000
M1
5n2 + 145n − 12 000 ≥ 0, or n2 + 29n − 2400 ≥ 0
A1
critical value n = 36.59…
M1
37 instalments
A1
Exam tip: After finding the critical value 36.6, check both neighbours: S36 = 11 700 is not enough, S37 = 12 210 is. That confirms 37.
An arithmetic series has first term a and common difference d.
(a)
Prove that the sum of the first n terms of the series is Sn = ½n[2a + (n − 1)d].
(4)
(b)
For a different arithmetic series, the sum of the first n terms is 3n2 + 2n for every positive integer n. Find the first term and the common difference of this series.
(3)
(c)
Find which term of this series is equal to 101.
(2)
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Answer
Mark
Notes
Part (a)
Sn = a + (a + d) + … + [a + (n − 1)d], with first terms and last term shown
B1
Sn = [a + (n − 1)d] + [a + (n − 2)d] + … + a, the same terms reversed
M1
adds: 2Sn = n[2a + (n − 1)d], with each of the n pairs equal to 2a + (n − 1)d
M1
divides by 2 to reach the given result, with a conclusion
A1
Part (b)
first term = S1 = 3 + 2 = 5
B1
S2 = 16, so the second term is 16 − 5 = 11
M1
d = 11 − 5 = 6
A1
Part (c)
5 + 6(n − 1) = 101
M1
n = 17, the 17th term
A1
Common error: Taking 3n2 + 2n to be the nth term. It is the sum; a single term is a difference of sums, un = Sn − Sn−1.
A sequence is defined by un+1 = −1/(un + 1), u1 = 1.
(a)
Find u2, u3 and u4.
(2)
(b)
State the order (period) of the sequence.
(1)
(c)
Find Σr=150ur.
(3)
(d)
Find u100.
(1)
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Answer
Mark
Notes
Part (a)
u2 = −½
B1
u3 = −2 and u4 = 1
B1
Part (b)
u4 = u1, so the sequence is periodic with order 3 ft
B1
Part (c)
sum of one cycle = 1 − ½ − 2 = −1.5
B1
50 = 16 × 3 + 2: 16 cycles plus the two terms 1 and −½
M1
16 × (−1.5) + 1 − ½ = −23.5
A1
Part (d)
100 = 33 × 3 + 1, so u100 = u1 = 1
B1
Exam tip: For a periodic sequence, find how many complete cycles fit and how many terms are left over: n = (number of cycles) × (period) + (remainder).
5/(n + 3) > 0 for every n ≥ 1, so every term is less than 2
B1
Part (e)
5/(n + 3) < 0.05, so n + 3 > 100
M1
n > 97, so the 98th term
A1
Exam tip: “Hence” means use the result just shown: the sign of the difference of consecutive terms decides whether a sequence is increasing or decreasing, for every n at once.
The points (n, un) representing the terms of an arithmetic sequence u1, u2, u3, … all lie on the dashed line in Figure 1. The labelled points P and Q represent the 3rd term and the 10th term.
(a)
Find the common difference and the first term of the sequence.
(3)
(b)
Find Σr=125ur.
(2)
(c)
Find how many terms of the sequence lie between 100 and 250.
(2)
(d)
Find the sum of the terms that lie between 100 and 250.
(2)
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Answer
Mark
Notes
Part (a)
d = (32 − 11)/(10 − 3), the gradient of the line
M1
d = 3
A1
u1 = 11 − 2 × 3 = 5
A1
Part (b)
½ × 25 × [2 × 5 + 24 × 3]
M1
= 1025
A1
Part (c)
un = 3n + 2: un > 100 gives n ≥ 33 (u33 = 101) and un < 250 gives n ≤ 82 (u82 = 248)
M1
82 − 33 + 1 = 50 terms
A1
Part (d)
½ × 50 × (101 + 248)
M1
= 8725 ft on their number of terms
A1
Exam tip: For an arithmetic sequence the points (n, un) lie on a straight line, and its gradient is the common difference. Counting terms from term 33 to term 82 inclusive gives 82 − 33 + 1, not 82 − 33.
Figure 1 shows a square S1 of side 20 cm. The vertices of square S2 are the midpoints of the sides of S1, the vertices of S3 are the midpoints of the sides of S2, and so on, forming an infinite sequence of squares S1, S2, S3, …
(a)
Show that the side lengths of the squares form a geometric sequence with common ratio 1/√2.
(2)
(b)
Find the exact area of S8.
(2)
(c)
Find the sum of the perimeters of all the squares, giving your answer in the form a + b√2, where a and b are integers.
(3)
(d)
Find the number of the first square in the sequence whose area is less than 0.01 cm2.
(3)
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Answer
Mark
Notes
Part (a)
a side of the next square is the hypotenuse of a right-angled triangle whose shorter sides are each half a side s of the square before
M1
√(¼s2 + ¼s2) = s/√2, so every side is 1/√2 times the side before
A1
Part (b)
areas form a geometric sequence with first term 400 and ratio ½: 400 × (½)7
M1
= 25/8 cm2
A1
Part (c)
S∞ = 80/(1 − 1/√2)
M1
= 80√2/(√2 − 1), then multiplies by (√2 + 1)/(√2 + 1)
M1
= 160 + 80√2 cm
A1
Part (d)
400 × 0.5n−1 < 0.01, so 0.5n−1 < 1/40 000
M1
n − 1 > log 40 000 / log 2 = 15.3 (3 s.f.)
M1
n = 17: S17
A1
Exam tip: If the sides are multiplied by 1/√2 each time, the areas are multiplied by (1/√2)2 = ½. Using the area ratio avoids surds in part (b).
Figure 1 shows the straight line y = f(x), which passes through the points A(0, 9) and B(10, 4), together with the line y = x. A sequence is defined by xn+1 = f(xn), x1 = 2, and the cobweb diagram on Figure 1 starts at x1.
(a)
Use Figure 1 to show that f(x) = 9 − ½x.
(2)
(b)
Find x2, x3 and x4.
(2)
(c)
State, with a reason, whether the sequence is increasing, decreasing or neither.
(1)
(d)
The sequence converges to a limit L. Find L.
(2)
(e)
By considering en = xn − 6, or otherwise, find the smallest value of n for which |xn − 6| < 0.001.
(3)
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Answer
Mark
Notes
Part (a)
gradient = (4 − 9)/(10 − 0) = −½
M1
y-intercept 9, so f(x) = 9 − ½x
A1
Part (b)
x2 = 8, x3 = 5
B1
x4 = 6.5
B1
Part (c)
neither: the terms 2, 8, 5, 6.5, … go up and down alternately, as the cobweb spirals
Exam tip: A cobweb that spirals in means the terms land alternately above and below the limit; a staircase means they approach it from one side. A line with negative gradient gives the spiral.
Figure 1 shows the first five rows of a triangular array of the positive integers. Row n contains n consecutive integers, and each row starts with the integer after the last number of the row before.
(a)
Explain why the last number in row n is ½n(n + 1).
(2)
(b)
Show that the first number in row n is ½(n2 − n + 2).
(2)
(c)
Find the number of the row that contains the integer 500.
(2)
(d)
Show that the sum of the numbers in row n is ½n(n2 + 1).
(2)
(e)
Hence find the sum of the numbers in row 20.
(2)
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Answer
Mark
Notes
Part (a)
the last number in row n equals how many numbers there are in rows 1 to n, which is 1 + 2 + 3 + … + n
M1
the sum of the first n natural numbers is ½n(n + 1)
A1
Part (b)
one more than the last number in row n − 1: ½(n − 1)n + 1
M1
= ½(n2 − n + 2)
A1
Part (c)
last number in row 31 is 496 and in row 32 is 528
M1
row 32
A1
Part (d)
row n is an arithmetic series of n terms, so its sum is ½n × (first + last)
M1
first + last = ½(n2 − n + 2) + ½(n2 + n) = n2 + 1, giving ½n(n2 + 1)
A1
Part (e)
½ × 20 × (400 + 1)
M1
= 4010
A1
Common error: Giving row 31 in part (c). The number 496 ends row 31, so 500 is in the next row; check where the row boundaries fall on each side.
Ana and Ben each start a savings plan in the same month. Both save $200 in month 1. Ana's plan: each month she saves $15 more than in the month before. Ben's plan: each month he saves 4% more than in the month before.
(a)
Find the amount each of them saves in month 12. Give Ben's amount to the nearest dollar.
(3)
(b)
Find the total amount each of them has saved after 24 months. Give Ben's total to the nearest dollar.
(4)
(c)
Find the number of the month in which Ben's total savings first exceed $10 000.
(3)
(d)
Ana's total savings first exceed $10 000 in month N. Find N, and state who reaches a total of more than $10 000 first.
Exam tip: The critical value 28.01 is very close to a whole number, so check it: Ben's total after 28 months is about $9994, just short, and after 29 months it is $10 593.
Exam tip: Both values of r satisfy |r| < 1, so both give a sum to infinity of 27. The extra information (all terms positive) is what picks out r = 2/3.
A patient takes a 20 mg dose of a medicine at the same time every day. A model assumes that, by the time of the next dose, 30% of the amount of medicine in the body remains. Let An mg be the amount in the body just after the nth dose, so that A1 = 20 and An+1 = 0.3An + 20.
(a)
Find A2 and A3.
(2)
(b)
Explain why An = 20 + 20(0.3) + 20(0.3)2 + … + 20(0.3)n−1, and hence show that An = (200/7)(1 − 0.3n).
(3)
(c)
Show that the sequence A1, A2, A3, … is increasing.
(2)
(d)
State, to 3 significant figures, the amount that the model predicts in the long term just after a dose, giving a reason.
(2)
(e)
Find the number of the first dose after which the amount in the body exceeds 28.5 mg.
(3)
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Answer
Mark
Notes
Part (a)
A2 = 0.3 × 20 + 20 = 26
B1
A3 = 0.3 × 26 + 20 = 27.8
B1
Part (b)
the latest dose contributes 20, the dose one day earlier 20 × 0.3, the dose j days earlier 20 × 0.3j, back to the first dose
B1
geometric series with a = 20, r = 0.3, n terms: 20(1 − 0.3n)/(1 − 0.3)
M1
= (200/7)(1 − 0.3n)
A1
Part (c)
An+1 − An = (200/7)(0.3n − 0.3n+1) = 20 × 0.3n
M1
which is positive for every n, so the sequence is increasing
Exam tip: A recurrence such as An+1 = 0.3An + 20 is not itself geometric, but writing out what each earlier dose still contributes turns it into a geometric series you can sum.
Figure 1 shows the start of a path drawn from the origin O. The first segment, OP1, has length 8 units along the positive x-axis. At the end of each segment the path turns 90° anticlockwise, and each segment is half as long as the segment before it. The path continues indefinitely through P1, P2, P3, P4, …
(a)
Find the total length of the first 6 segments.
(2)
(b)
Show that the total length of the path can never be 16 units or more.
(2)
(c)
Find the least number of segments for which the total length of the path exceeds 15.99 units.
(3)
(d)
By considering the horizontal segments and the vertical segments separately, show that the path approaches the point (6.4, 3.2).
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
8(1 − 0.56)/(1 − 0.5)
M1
= 15.75 units
A1
Part (b)
S∞ = 8/(1 − ½) = 16
B1
Sn = 16(1 − 0.5n) and 0.5n > 0, so every total is less than 16
B1
Part (c)
16(1 − 0.5n) > 15.99, so 0.5n < 1/1600
M1
n > log 1600 / log 2 = 10.6 (3 s.f.)
M1
11 segments
A1
Part (d)
the x-moves 8, −2, 0.5, … form a geometric series with common ratio −¼
M1
x → 8/(1 + ¼) = 6.4
A1
the y-moves 4, −1, ¼, … form a geometric series with common ratio −¼
M1
y → 4/(1 + ¼) = 3.2, so the path approaches (6.4, 3.2)
A1
Common error: Using ½ or ¼ as the ratio for the horizontal moves. Two quarter-turns reverse the direction, so consecutive horizontal moves have opposite signs and the ratio is −¼.
Figure 1 shows the centre of a circular target. Region 1 is a disc of radius 2 cm. Each further region is a ring 2 cm wide, so region k has outer radius 2k cm. A complete target has 20 regions.
(a)
Show that the area of region k is 4π(2k − 1) cm2.
(3)
(b)
Hence show that the areas of regions 1, 2, 3, … form an arithmetic sequence, stating its first term and common difference.
(2)
(c)
Use the formula for the sum of an arithmetic series to show that the total area of regions 1 to n is 4πn2 cm2, and explain how Figure 1 confirms this result.
(3)
(d)
In the complete target the odd-numbered regions are gold. Find the exact total area that is gold.
(3)
(e)
A designer wants a larger target of the same type with a total area of more than 8000 cm2. Find the least number of regions it needs.
(2)
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Answer
Mark
Notes
Part (a)
outer radius 2k and inner radius 2(k − 1)
B1
π(2k)2 − π(2k − 2)2
M1
= π(8k − 4) = 4π(2k − 1)
A1
Part (b)
4π(2k + 1) − 4π(2k − 1) = 8π, which is constant
B1
first term 4π, common difference 8π
B1
Part (c)
½n[8π + 8π(n − 1)]
M1
= 4πn2
A1
regions 1 to n together form a disc of radius 2n, whose area is π(2n)2 = 4πn2
B1
Part (d)
regions 1, 3, 5, …, 19: 10 regions with areas 4π, 20π, 36π, …, an arithmetic series with common difference 16π
M1
½ × 10 × [8π + 9 × 16π]
M1
= 760π cm2
A1
Part (e)
4πn2 > 8000, so n > √(2000/π) = 25.2
M1
26 regions
A1
Common error: Taking the area of region k as π(2k)2, the whole disc out to radius 2k. A ring's area is the outer disc minus the inner disc.
Figure 1 shows the first three stages of a branching design. Stage 1 is a trunk of length 50 cm. At each later stage, three new branches grow from the end of every branch added at the stage before, and each new branch is 0.3 times as long as the branch it grows from. The design continues indefinitely.
(a)
Find the total length of the branches added at stage 3.
(2)
(b)
Show that the total length of the branches added at stage n is 50(0.9)n−1 cm.
(2)
(c)
Find the total length of the design after 10 stages, giving your answer to 3 significant figures.
(2)
(d)
Explain why the total length of the design can never reach 5 m.
(2)
(e)
Find the first stage after which the total length of the design exceeds 4.5 m.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
9 branches, each 50 × 0.32 = 4.5 cm long
M1
= 40.5 cm
A1
Part (b)
3n−1 branches, each of length 50(0.3)n−1 cm
M1
product = 50(3 × 0.3)n−1 = 50(0.9)n−1
A1
Part (c)
50(1 − 0.910)/(1 − 0.9)
M1
= 325.6… = 326 cm
A1
Part (d)
0 < 0.9 < 1, so S∞ = 50/(1 − 0.9) = 500 cm = 5 m
B1
Sn = 500(1 − 0.9n) < 500 for every n, since 0.9n > 0
B1
Part (e)
500(1 − 0.9n) > 450, so 0.9n < 0.1
M1
n > log 0.1 / log 0.9 = 21.9 (3 s.f.), reversing the inequality when dividing by log 0.9 < 0
M1
stage 22
A1
Exam tip: The number of branches is multiplied by 3 at each stage while each length is multiplied by 0.3. Multiply the two ratios to get the ratio for the length added per stage: 3 × 0.3 = 0.9.
A small quarry starts production. Figure 1 shows its output, in tonnes, in each of its first 24 months. In phase 1 (months 1 to 12) the monthly outputs form an arithmetic sequence. In phase 2 (from month 13 onwards) the output each month is 95% of the output in the month before.
(a)
Use Figure 1 to find the common difference in phase 1, and show that the total output in phase 1 is 3720 tonnes.
(3)
(b)
Find the output in month 20, giving your answer to 3 significant figures.
(2)
(c)
Find the total output from month 13 to month 36 inclusive, giving your answer to 3 significant figures.
(3)
(d)
Find the first month in which the output is less than 100 tonnes.
(3)
(e)
Show that, if the model continued indefinitely, the total output could never exceed 11 700 tonnes.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
d = (420 − 200)/11
M1
= 20 tonnes
A1
½ × 12 × (200 + 420) = 3720
A1
Part (b)
420 × 0.958
M1
= 278.6… = 279 tonnes
A1
Part (c)
geometric series with first term 420 × 0.95 = 399, common ratio 0.95 and 24 terms
M1
399(1 − 0.9524)/(1 − 0.95)
M1
= 5649.9… = 5650 tonnes
A1
Part (d)
420 × 0.95k < 100, where k is the number of months after month 12
M1
k > log(100/420) / log 0.95 = 27.98
M1
k = 28, so month 40
A1
Part (e)
phase 2 sum to infinity = 399/(1 − 0.95) = 7980
M1
3720 + 7980 = 11 700, and every finite total is less than this because all terms are positive
A1
Exam tip: The critical value 27.98 is very close to 28, so check both neighbours: month 39 has output about 105.1 tonnes and month 40 about 99.9 tonnes.
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