Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 1: Sequences and series worked solutions

Worked solutions to the practice questions in Chapter 1 of Photon Guidebook: Mathematics Analysis and Approaches SL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

  • Chapter 1 of 12
  • 23 practice questions
  • 4 Going further
  • Free, no sign-up
Start with Basics All 12 chapters

Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[5 marks]
A sequence is defined by u_n=31-4n, for n\ge 1.
(a) Find u_1, u_2 and u_3.
[2]
(b) Show that u_{n+1}-u_n is the same for every n, and state the common difference.
[2]
(c) Find u_{15}.
[1]
Show solutionHide solution
(a) u_1=27,\ u_2=23,\ u_3=19
A2
Note: award A1 for two correct values.
(b) u_{n+1}-u_n=\left(31-4(n+1)\right)-(31-4n)
M1
=-4, which does not depend on n, so d=-4
A1
(c) u_{15}=31-60=-29
A1

Final answer

(a) 27,23,19 (b) d=-4 (c) -29

Question 2

[6 marks]
Write each sum out in full and find its value.
(a) \displaystyle\sum_{r=1}^{4}(5r-1)
[2]
(b) \displaystyle\sum_{k=3}^{6}(10-2k)
[2]
(c) \displaystyle\sum_{n=1}^{5}3
[2]
Show solutionHide solution
(a) 4+9+14+19
M1
=46
A1
(b) 4+2+0+(-2)
M1
=4
A1
(c) 3+3+3+3+3
M1
=15
A1
Note: in (b) award M0 if the sum starts at k=1.

Final answer

(a) 46 (b) 4 (c) 15

Question 3

[6 marks]
An arithmetic sequence has first term \dfrac{1}{3} and common difference \dfrac{2}{3}.
(a) Show that u_n=\dfrac{2n-1}{3}.
[2]
(b) Find the sum of the first 20 terms, giving your answer as a fraction.
[2]
(c) Find which term of the sequence is equal to 27.
[2]
Show solutionHide solution
(a) u_n=\dfrac{1}{3}+\dfrac{2}{3}(n-1)
M1
=\dfrac{1+2n-2}{3}
A1
=\dfrac{2n-1}{3}
AG
(b) u_{20}=\dfrac{39}{3}=13, S_{20}=\dfrac{20}{2}\left(\dfrac{1}{3}+13\right)
M1
=\dfrac{400}{3}
A1
(c) \dfrac{2n-1}{3}=27 \Rightarrow 2n-1=81
M1
n=41, the 41st term
A1

Final answer

(a) given result (b) 400/3 (c) 41st

Question 4

[6 marks]
The fourth term of an arithmetic sequence is 23 and the eleventh term is 2.
(a) Determine d and u_1.
[4]
(b) Calculate S_{11}.
[2]
Show solutionHide solution
(a) u_1+3d=23 and u_1+10d=2
M1
7d=-21 \Rightarrow d=-3
A1
u_1=23-3(-3)
M1
u_1=32
A1
(b) S_{11}=\dfrac{11}{2}(32+2)
M1
=187
A1

Final answer

(a) d=-3, u_1=32 (b) 187

Question 5

[3 marks]
Consider the geometric sequence 243,\ -81,\ 27,\ \dots
(a) Find the common ratio.
[1]
(b) Find the exact value of the seventh term.
[2]
Show solutionHide solution
(a) r=\dfrac{-81}{243}=-\dfrac{1}{3}
A1
(b) u_7=243\left(-\dfrac{1}{3}\right)^{6}
M1
=\dfrac{243}{729}=\dfrac{1}{3}
A1

Final answer

(a) -1/3 (b) 1/3

Question 6

[6 marks]
The second term of a geometric sequence is -48 and the fifth term is 162.
(a) Find the common ratio.
[3]
(b) Find the first term and the sixth term.
[3]
Show solutionHide solution
(a) u_1r=-48,\ u_1r^{4}=162
M1
r^{3}=\dfrac{162}{-48}=-\dfrac{27}{8}
A1
r=-\dfrac{3}{2}
A1
(b) u_1=\dfrac{-48}{-3/2}=32
A1
u_6=162\times\left(-\dfrac{3}{2}\right)
M1
=-243
A1

Final answer

(a) -3/2 (b) u_1=32, u_6=-243

Question 7

[4 marks]
For an infinite geometric series with first term u_1 and common ratio r, where |r|<1,
S_\infty=\dfrac{u_1}{1-r}.
Find the sum to infinity of each series.
(a) 120+40+\dfrac{40}{3}+\dots
[2]
(b) 50-20+8-\dots
[2]
Show solutionHide solution
(a) r=\dfrac{1}{3}, S_\infty=\dfrac{120}{1-\frac{1}{3}}
M1
S_\infty=180
A1
(b) r=-\dfrac{2}{5}, S_\infty=\dfrac{50}{1-\left(-\frac{2}{5}\right)}=\dfrac{50}{\frac{7}{5}}
M1
S_\infty=\dfrac{250}{7}\ \left(=35\tfrac{5}{7}\right)
A1

Final answer

(a) 180 (b) 250/7

Question 8

[8 marks]
Consider the infinite series \displaystyle\sum_{k=0}^{\infty} 6(2x-3)^k.
(a) Write down the common ratio of the series.
[1]
(b) Determine the interval of convergence.
[3]
(c) Show that, for these values of x, the sum of the series is \dfrac{3}{2-x}.
[2]
(d) Solve for x when the infinite total equals 12.
[2]
Show solutionHide solution
(a) r=2x-3
A1
(b) |2x-3|<1 \Rightarrow -1<2x-3<1
M1
2<2x<4 (A1)
1<x<2
A1
(c) first term 6, \ S_\infty=\dfrac{6}{1-(2x-3)}
M1
=\dfrac{6}{4-2x}
A1
=\dfrac{3}{2-x}
AG
(d) \dfrac{3}{2-x}=12 \Rightarrow 2-x=\dfrac{1}{4}
M1
x=\dfrac{7}{4}, which lies in 1<x<2
A1

Final answer

(a) 2x-3 (b) 1<x<2 (c) given result (d) 7/4

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[5 marks]
The numbers \log_3 2, \ \log_3(x+1) and \log_3(x+4) are, in this order, three consecutive terms of an arithmetic sequence, where x > -1.
Find the exact value of x.
[5]
Show solutionHide solution
Arithmetic: 2\log_3(x+1) = \log_3 2 + \log_3(x+4).
[1]
Laws of logarithms: \log_3(x+1)^2 = \log_3\bigl(2(x+4)\bigr), so (x+1)^2 = 2x + 8.
[1]
x^2 + 2x + 1 = 2x + 8 \Rightarrow x^2 = 7.
[1]
x = \pm\sqrt7.
[1]
The condition x > -1 rejects -\sqrt7 \approx -2.65, so x = \sqrt7.
[1]

Final answer

x=\sqrt7

Question 10

[7 marks]
An arithmetic sequence has u_2 + u_6 = 24 and u_{10} = 4u_1.
(a) Determine u_1 and d.
[5]
(b) Find the sum of the first 12 terms.
[2]
Show solutionHide solution
(a) u_2 + u_6 = (u_1 + d) + (u_1 + 5d) = 2u_1 + 6d = 24.
[2]
u_{10} = 4u_1 gives u_1 + 9d = 4u_1, so u_1 = 3d.
[1]
Substitute: 6d + 6d = 24 \Rightarrow d = 2.
[1]
d = 2, u_1 = 6.
[1]
(b) S_{12} = \dfrac{12}{2}\bigl(2(6) + 11(2)\bigr)
[1]
= 6 \times 34 = 204.
[1]

Final answer

(a) u_1=6, d=2 (b) 204

Question 11

[7 marks]
An arithmetic sequence has first term u_1 = 4 and common difference d = 7. A new sequence v_1,\ v_2,\ v_3,\ \ldots is formed by taking every third term of (u_n), so that v_1 = u_1, \ v_2 = u_4, \ v_3 = u_7, and so on.
(a) Show that (v_n) is an arithmetic sequence and write down its common difference.
[3]
(b) Find an expression for v_n in terms of n.
[2]
(c) Find the value of n for which v_n = 214.
[2]
Show solutionHide solution
(a) v_n = u_{3n-2} and v_{n+1} = u_{3n+1}.
[1]
v_{n+1} - v_n = u_{3n+1} - u_{3n-2} = 3d = 21, which does not depend on n.
[1]
So (v_n) is arithmetic with common difference 21.
[1]
(b) v_1 = 4, so v_n = 4 + 21(n-1)
[1]
= 21n - 17.
[1]
(c) 21n - 17 = 214 \Rightarrow 21n = 231
[1]
n = 11.
[1]

Final answer

(a) d_v=21 (b) v_n=21n-17 (c) 11

Question 12

[6 marks]
An arithmetic sequence has first term 61 and common difference -2.8.
(a) Find the first term of the sequence that is negative.
[3]
Write S_n for the total through term n.
(b) Find the greatest value of S_n, and state the value of n for which it occurs.
[3]
Show solutionHide solution
(a) u_n = 61 - 2.8(n-1) < 0 \Rightarrow n - 1 > 21.78\ldots
[2]
GDC table: u_{22} = 2.2 > 0 and u_{23} = -0.6 < 0. The first negative term is u_{23} = -0.6.
[1]
(b) S_n is greatest when every positive term has been added and no negative term has, so n = 22.
[1]
S_{22} = \dfrac{22}{2}\bigl(2(61) + 21(-2.8)\bigr)
[1]
= 11 \times 63.2 = 695.2 \approx 695 (3 s.f.).
[1]

Final answer

(a) 23rd term, -0.600 (b) 695 at n=22

Question 13

[6 marks]
A geometric sequence has u_2 = 6 and u_4 = 24.
(a) Show that the common ratio is either 2 or -2.
[2]
It is also given that u_3 < 0.
(b) Find the common ratio and the first term.
[2]
(c) Find u_6.
[2]
Show solutionHide solution
(a) u_4 = u_2 r^2, so r^2 = \dfrac{24}{6} = 4.
[1]
r = 2 or r = -2.
[1]
(b) u_3 = u_2 r = 6r < 0 requires r < 0, so r = -2.
[1]
u_1 = \dfrac{u_2}{r} = \dfrac{6}{-2} = -3.
[1]
(c) u_6 = u_2 r^4 = 6 \times 16
[1]
= 96.
[1]

Final answer

(a) given result (b) r=-2, u_1=-3 (c) 96

Question 14

[7 marks]
The expressions k + 1, \ k - 1 and 2k - 5 begin a geometric sequence in the order given.
(a) Show that k^2 - k - 6 = 0.
[2]
The corresponding geometric series has a sum to infinity.
(b) Find the value of k.
[3]
(c) Find the sum to infinity.
[2]
Show solutionHide solution
(a) For a geometric sequence the middle term squared equals the product of its neighbours: (k-1)^2 = (k+1)(2k-5).
[1]
k^2 - 2k + 1 = 2k^2 - 3k - 5.
[1]
0 = k^2 - k - 6, i.e. k^2 - k - 6 = 0.
AG
(b) (k - 3)(k + 2) = 0, so k = 3 or k = -2.
[2]
k = -2 gives the terms -1,\ -3,\ -9 with r = 3, and |r| > 1 so there is no sum to infinity. Hence k = 3.
[1]
(c) k = 3: the terms are 4,\ 2,\ 1, so r = \frac12.
[1]
S_\infty = \dfrac{4}{1 - \frac12} = 8.
[1]

Final answer

(a) given result (b) 3 (c) 8

Question 15

[5 marks]
(a) Find the sum of the first 10 terms of the series 4 + 4 + 4 + \cdots, and explain why the formula \dfrac{u_1(r^n - 1)}{r - 1} cannot be used.
[2]
Consider the series 7 + 7x + 7x^2 + \cdots + 7x^9, where x \neq 1.
(b) Write down a simplified expression for the sum in terms of x.
[1]
(c) Use your answer to (b) to find the sum when x = -1, and verify this result by grouping the terms in pairs.
[2]
Show solutionHide solution
(a) The series is 10 terms each equal to 4: the sum is 40.
[1]
Here r = 1, so r - 1 = 0 and the formula would divide by zero.
[1]
(b) u_1 = 7, r = x, n = 10: sum = \dfrac{7(x^{10} - 1)}{x - 1}.
[1]
(c) When x = -1: \dfrac{7\bigl((-1)^{10} - 1\bigr)}{-2} = 0.
[1]
Check: 7 - 7 + 7 - 7 + \cdots consists of 5 pairs, each summing to 0.
[1]

Final answer

(a) 40; r=1 (b) 7(x^{10}-1)/(x-1) (c) 0

Question 16

[6 marks]
A geometric series has first term 4 and common ratio 1 - 3x.
(a) Determine the convergence interval for x.
[3]
(b) Write down S_\infty in terms of x, in its simplest form.
[1]
(c) Given that S_\infty = 10, find the value of x.
[2]
Show solutionHide solution
(a) The sum to infinity exists when |1 - 3x| < 1.
[1]
-1 < 1 - 3x < 1 \Rightarrow -2 < -3x < 0.
[1]
0 < x < \dfrac23.
[1]
(b) S_\infty = \dfrac{4}{1 - (1 - 3x)} = \dfrac{4}{3x}.
[1]
(c) \dfrac{4}{3x} = 10 \Rightarrow x = \dfrac{4}{30}
[1]
= \dfrac{2}{15}, which lies in 0 < x < \frac23.
[1]

Final answer

(a) 0<x<2/3 (b) 4/(3x) (c) 2/15

Question 17

[7 marks]
A sequence has partial totals S_n = 4n^2 - n.
(a) Find u_1.
[1]
(b) Show that u_n = 8n - 5 for n \geq 2.
[3]
(c) Hence show that the sequence is arithmetic, and write down the common difference.
[2]
(d) Find the value of k for which u_k = 107.
[1]
Show solutionHide solution
(a) u_1 = S_1 = 4(1) - 1 = 3.
[1]
(b) For n \geq 2: u_n = S_n - S_{n-1}.
[1]
= (4n^2 - n) - \bigl(4(n-1)^2 - (n-1)\bigr)
[1]
= 4n^2 - n - 4n^2 + 8n - 4 + n - 1
[1]
= 8n - 5.
AG
(c) u_1 = 3 = 8(1) - 5, so the formula holds for n = 1 as well, and u_{n+1} - u_n = 8 for all n.
[1]
The sequence is arithmetic with d = 8.
[1]
(d) 8k - 5 = 107 \Rightarrow k = 14.
[1]

Final answer

(a) 3 (b) given result (c) d=8 (d) 14

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 18

[8 marks]
The sequence (u_n) has partial totals S_n = n^2 + 5n.
(a) Find an expression for u_n.
[3]
(b) Show that u_1,\ u_4,\ u_{10} begin a geometric sequence in that order.
[2]
The geometric sequence in part (b) continues u_1,\ u_4,\ u_{10},\ u_{22},\ \ldots
(c) The 8th term of the geometric sequence is u_k. Find the value of k.
[3]
Show solutionHide solution
(a) u_1 = S_1 = 6.
[1]
For n \geq 2: u_n = S_n - S_{n-1} = n^2 + 5n - \bigl((n-1)^2 + 5(n-1)\bigr) = 2n + 4.
[1]
This also gives u_1 = 6, so u_n = 2n + 4 for all n \geq 1.
[1]
(b) u_1 = 6, u_4 = 12, u_{10} = 24.
[1]
\dfrac{12}{6} = \dfrac{24}{12} = 2, a constant ratio, so they are geometric.
[1]
(c) The geometric sequence has first term 6 and ratio 2, so its 8th term is 6 \times 2^7 = 768.
[2]
2k + 4 = 768 \Rightarrow k = 382.
[1]

Final answer

(a) 2n+4 (b) 6,12,24; ratio 2 (c) 382

Question 19

[6 marks]
An arithmetic sequence (a_n) has first term 1 and common difference 2. A new sequence (b_n) is defined by b_n = 3^{a_n}.
(a) Show that (b_n) is a geometric sequence, and state its first term and common ratio.
[3]
(b) Find \displaystyle\sum_{n=1}^{5} b_n.
[3]
Show solutionHide solution
(a) a_n = 1 + 2(n-1) = 2n - 1, so b_n = 3^{2n-1}.
[1]
\dfrac{b_{n+1}}{b_n} = \dfrac{3^{2n+1}}{3^{2n-1}} = 3^2 = 9, a constant, so (b_n) is geometric.
[1]
First term b_1 = 3^1 = 3 and common ratio 9.
[1]
(b) S_5 = \dfrac{3(9^5 - 1)}{9 - 1}
[2]
= \dfrac{3 \times 59\,048}{8} = 22\,143.
[1]

Final answer

(a) b_1=3, r=9 (b) 22 143

Question 20

[14 marks]
Two towns collect recycling. Town A collects 1200 tonnes in month 1, and each month it collects 2\% more than in the previous month. Town B collects 1150 tonnes in month 1, and each month it collects 40 tonnes more than in the previous month.
(a) Find the amount collected in month 12 by each town, giving your answers to three significant figures.
[3]
(b) Show that the total amount collected by Town B in its first n months is 20n^2 + 1130n tonnes.
[3]
(c) Find the least number of months for which the total amount collected by Town A exceeds 80\,000 tonnes.
[3]
Each year consists of 12 months. Let Y_m be the total amount, in tonnes, collected by Town A during year m.
(d) Show that Y_1, Y_2, Y_3, \ldots form a geometric sequence, and find its common ratio to three significant figures.
[3]
(e) Find the total amount collected by Town A in its first 5 years, to the nearest tonne.
[2]
Show solutionHide solution
(a) Town A is geometric (u_1 = 1200, r = 1.02): A_{12} = 1200(1.02)^{11}
[1]
= 1492.05\ldots \approx 1490 tonnes.
[1]
Town B is arithmetic (u_1 = 1150, d = 40): B_{12} = 1150 + 11(40) = 1590 tonnes.
[1]
(b) Town B is arithmetic with u_1 = 1150 and d = 40: S_n = \dfrac n2\bigl(2(1150) + (n-1)40\bigr).
[1]
= \dfrac n2(2300 + 40n - 40)
[1]
= \dfrac n2(2260 + 40n)
[1]
= 1130n + 20n^2, i.e. 20n^2 + 1130n tonnes.
AG
(Check: n = 24 gives 20(576) + 1130(24) = 38\,640.)
(c) Town A is geometric with u_1 = 1200 and r = 1.02: S_n = \dfrac{1200(1.02^n - 1)}{0.02} = 60\,000(1.02^n - 1).
[1]
Solve 60\,000(1.02^n - 1) > 80\,000, i.e. 1.02^n > 2.333\ldots
GDC table: n = 42: S_n = 77\,835 < 80\,000 and n = 43: S_n = 80\,591 > 80\,000.
[1]
Least number of months: 43.
[1]
(d) Year m consists of months 12(m-1) + 1 to 12m, so Y_m = 1.02^{12(m-1)} \times Y_1, because every month's amount in year m is 1.02^{12(m-1)} times the corresponding month in year 1.
[1]
Hence \dfrac{Y_{m+1}}{Y_m} = 1.02^{12}, a constant, so (Y_m) is geometric.
[1]
Common ratio = 1.02^{12} = 1.2682\ldots \approx 1.27 (3 s.f.).
[1]
(e) The first 5 years are 60 months: S_{60} = 60\,000(1.02^{60} - 1)
[1]
= 136\,861.8\ldots \approx 136\,862 tonnes.
[1]
(Equivalently, Y_1 = 16\,094.5\ldots and the sum of five terms of a geometric sequence with r = 1.2682\ldots gives the same total.)

Final answer

(a) 1490, 1590 tonnes (b) given result (c) 43 months (d) 1.27 (e) 136 862 tonnes

Question 21

[12 marks]
A geometric series has first term a and common ratio r, where 0 < r < 1. Its sum to infinity is 40, and the sum of all its terms from the third term onwards is 10.
(a) Show that the sum of all the terms from the kth term onwards is \dfrac{u_k}{1 - r}.
[2]
(b) Determine r and a.
[4]
(c) Find the least value of k for which the sum of all the terms from the kth term onwards is less than 0.001.
[3]
(d) Determine the smallest positive integer n satisfying S_n>39.9.
[3]
Show solutionHide solution
(a) The terms from the kth onwards are u_k,\ u_k r,\ u_k r^2,\ \ldots, an infinite geometric series with first term u_k and ratio r.
[1]
Since |r| < 1 its sum is \dfrac{u_k}{1 - r}.
[1]
(b) \dfrac{a}{1 - r} = 40 and, by part (a) with k = 3, \dfrac{ar^2}{1 - r} = 10.
[1]
Dividing the second by the first: r^2 = \dfrac14.
[1]
0 < r < 1, so r = \dfrac12.
[1]
a = 40\left(1 - \frac12\right) = 20.
[1]
(c) u_k = 20\left(\tfrac12\right)^{k-1}, so the tail is \dfrac{u_k}{1 - r} = 40\left(\tfrac12\right)^{k-1} < 0.001.
[1]
GDC table: k = 16: 40/2^{15} = 0.00122 > 0.001; k = 17: 40/2^{16} = 0.00061 < 0.001.
[1]
Least k = 17.
[1]
(d) S_n = 40\left(1 - \left(\tfrac12\right)^n\right) > 39.9, so \left(\tfrac12\right)^n < 0.0025.
[1]
GDC table: n = 8: 0.00391 > 0.0025; n = 9: 0.00195 < 0.0025.
[1]
Least n = 9.
[1]

Final answer

(a) given result (b) r=1/2, a=20 (c) 17 (d) 9

Question 22

[13 marks]
(a) Show that \ln\left(1 + \dfrac1k\right) = \ln(k+1) - \ln k for k \in \mathbb{Z}^+.
[1]
(b) Hence find an expression, in terms of n, for \displaystyle\sum_{k=1}^{n}\ln\left(1 + \frac1k\right).
[3]
(c) Find the least value of n for which \displaystyle\sum_{k=1}^{n}\ln\left(1 + \frac1k\right) > \ln 50.
[2]
(d) Show that 1 - \dfrac1{k^2} = \dfrac{(k-1)(k+1)}{k^2}, and hence show that
\sum_{k=2}^{n}\ln\left(1 - \frac1{k^2}\right) = \ln\left(\frac{n+1}{2n}\right).
[5]
(e) State the value that \displaystyle\sum_{k=2}^{n}\ln\left(1 - \frac1{k^2}\right) approaches as n becomes very large.
[2]
Show solutionHide solution
(a) \ln\left(1 + \dfrac1k\right) = \ln\left(\dfrac{k+1}{k}\right) = \ln(k+1) - \ln k.
[1]
(b) Write out the terms: \bigl(\ln 2 - \ln 1\bigr) + \bigl(\ln 3 - \ln 2\bigr) + \cdots + \bigl(\ln(n+1) - \ln n\bigr).
[2]
Every term cancels except the first and last: the sum is \ln(n+1) - \ln 1 = \ln(n+1).
[1]
(c) \ln(n+1) > \ln 50 \Rightarrow n + 1 > 50.
[1]
Least n = 50.
[1]
(d) 1 - \dfrac1{k^2} = \dfrac{k^2 - 1}{k^2} = \dfrac{(k-1)(k+1)}{k^2}.
[1]
So \ln\left(1 - \dfrac1{k^2}\right) = \ln(k-1) + \ln(k+1) - 2\ln k = \bigl[\ln(k+1) - \ln k\bigr] - \bigl[\ln k - \ln(k-1)\bigr].
[2]
Summing from k = 2 to n, the first brackets telescope to \ln(n+1) - \ln 2 and the second brackets to \ln n - \ln 1 = \ln n.
[1]
Sum = \ln(n+1) - \ln 2 - \ln n = \ln\left(\dfrac{n+1}{2n}\right).
[1]
(e) As n becomes large, \dfrac{n+1}{2n} = \dfrac12 + \dfrac1{2n} \to \dfrac12.
[1]
The sum approaches \ln\dfrac12 = -\ln 2.
[1]

Final answer

(a) given result (b) \ln(n+1) (c) 50 (d) given result (e) -\ln2

Question 23

[17 marks]
A community gallery receives a grant in 24 monthly instalments. The first instalment is 180 dollars and each later instalment is 6% larger than the preceding one. Let T_n be the total received after n instalments.
(a) Calculate the sixth instalment.
[2]
(b) Show that T_n=3000(1.06^n-1).
[3]
(c) Find the first month in which the total received exceeds 4000 dollars.
[3]
(d) Find the complete grant T_{24}.
[2]
After month 24, 40% of the complete grant is assigned to conservation. Of this conservation fund, 20% is spent in year 1, and each later year's spending is 80% of the previous year's. Assume this schedule continues indefinitely.
(e) Explain why the fund is sufficient for all the scheduled spending.
[3]
(f) Find the least number of years by which more than 98% of the conservation fund has been spent.
[4]
Show solutionHide solution
(a) 180(1.06)^5 [M1] =241 dollars (3 s.f.) [A1].
(b) T_n=180(1.06^n-1)/(1.06-1) [M1]; 180/0.06=3000 [A1]; hence T_n=3000(1.06^n-1) [A1 AG].
(c) Solve 3000(1.06^n-1)>4000 [M1]. T_{14}=3782.7118\ldots<4000<T_{15}=4189.6745\ldots [R1]; month 15 [A1].
(d) T_{24}=3000(1.06^{24}-1) [M1] =9150 dollars (3 s.f.) [A1].
(e) Write F=0.4T_{24}, using the unrounded grant. Spending starts at 0.2F, ratio 0.8 [M1]; |0.8|<1, so a sum to infinity exists [R1]. Total 0.2F/(1-0.8)=F, exactly the fund [A1].
(f) Spending over k years is F(1-0.8^k) [M1]; require 0.8^k<0.02 [M1]. At 17 years the fraction spent is 0.977482\ldots, at 18 it is 0.981986\ldots [R1]. Least 18 years [A1].

Final answer

(a) 241 dollars (b) given result (c) month 15 (d) 9150 dollars (e) total spending equals the fund (f) 18 years

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Sums that collapse

G1 Worked example

[11 marks]
(a) Show that \dfrac1r-\dfrac1{r+2}=\dfrac{2}{r(r+2)}.
[1]
(b) Write the first four rows and the last two rows of \displaystyle\sum_{r=1}^{n}\left(\frac1r-\frac1{r+2}\right) one under another. Hence show that \displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)}=\frac34-\frac12\left(\frac1{n+1}+\frac1{n+2}\right).
[4]
(c) Write down the sum to infinity of \displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+2)}.
[1]
(d) Find the least n for which the partial sum in (b) is within 0.01 of this sum to infinity.
[3]
(e) Without writing out any rows, explain why \displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+3)}=\frac13\left(1+\frac12+\frac13\right), and evaluate it.
[2]
Show solutionHide solution
(a) \dfrac1r-\dfrac1{r+2}=\dfrac{(r+2)-r}{r(r+2)}=\dfrac{2}{r(r+2)}
A1AG
(b) \begin{array}{rcl}r=1:&1&-\frac13\\r=2:&\frac12&-\frac14\\r=3:&\frac13&-\frac15\\r=4:&\frac14&-\frac16\\&\vdots&\\r=n-1:&\frac1{n-1}&-\frac1{n+1}\\r=n:&\frac1n&-\frac1{n+2}\end{array}
M1
Each negative term reappears as a positive term two rows later, so only 1+\frac12-\frac1{n+1}-\frac1{n+2} survives.
A1
By (a), each term of the required sum is half a row, so halve the survivors
M1
\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)}=\frac12\cdot\frac32-\frac12\left(\frac1{n+1}+\frac1{n+2}\right)=\frac34-\frac12\left(\frac1{n+1}+\frac1{n+2}\right)
A1AG
(c) both fractions in the bracket tend to 0 as n\to\infty, so the sum to infinity is \dfrac34
A1
(d) \dfrac34-S_n=\dfrac12\left(\dfrac1{n+1}+\dfrac1{n+2}\right)<0.01\iff\dfrac1{n+1}+\dfrac1{n+2}<0.02
M1
n=98: \frac1{99}+\frac1{100}=0.02010\ldots>0.02; ; n=99: \frac1{100}+\frac1{101}=0.01990\ldots<0.02
R1
n=99
A1
Note: the gap from the limit is 0.01005\ldots at n=98 and 0.00995\ldots at n=99. Award R1 for both neighbouring values.
(e) \dfrac1r-\dfrac1{r+3}=\dfrac{3}{r(r+3)}, and in the rows each negative term returns three rows later, so the first three positive terms 1,\frac12,\frac13 survive while the last three negative terms tend to 0
M1
\displaystyle\sum_{r=1}^{\infty}\frac1{r(r+3)}=\frac13\left(1+\frac12+\frac13\right)=\frac{11}{18}
A1

Final answer

  • (c) \frac34
  • (d) n=99
  • (e) \frac{11}{18}

G1 Investigation

[13 marks]
For k\in\mathbb{Z}^+, let P_k=\displaystyle\sum_{r=1}^{\infty}\frac{1}{r(r+1)(r+2)\cdots(r+k)}, where each denominator is a product of k+1 consecutive integers.
(a) Use \dfrac{1}{r(r+1)}=\dfrac1r-\dfrac1{r+1} to find P_1.
[1]
(b) Show that \dfrac{1}{r(r+1)}-\dfrac{1}{(r+1)(r+2)}=\dfrac{2}{r(r+1)(r+2)}.
[2]
(c) Hence find \displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)} in terms of n, and deduce the value of P_2.
[3]
(d) Use the same method to find P_3. Using P_1, P_2 and P_3, conjecture a formula for P_k in terms of k.
[3]
(e) Prove your conjecture.
[4]
Show solutionHide solution
(a) \displaystyle\sum_{r=1}^{n}\left(\frac1r-\frac1{r+1}\right)=1-\frac1{n+1}
M1
\to1 as n\to\infty, so P_1=1 (A1 included with M1: 1 mark in total)
Note: award the mark for P_1=1 with the collapsed sum seen.
(b) \dfrac{1}{r(r+1)}-\dfrac{1}{(r+1)(r+2)}=\dfrac{(r+2)-r}{r(r+1)(r+2)}
M1
=\dfrac{2}{r(r+1)(r+2)}
A1AG
(c) Let g(r)=\dfrac{1}{r(r+1)}. Then \displaystyle\sum_{r=1}^{n}\frac{2}{r(r+1)(r+2)}=\sum_{r=1}^{n}\bigl(g(r)-g(r+1)\bigr)=g(1)-g(n+1)
M1
\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}=\frac12\left(\frac12-\frac{1}{(n+1)(n+2)}\right)=\frac14-\frac{1}{2(n+1)(n+2)}
A1
\dfrac{1}{2(n+1)(n+2)}\to0, so P_2=\dfrac14
A1
(d) \dfrac{1}{r(r+1)(r+2)}-\dfrac{1}{(r+1)(r+2)(r+3)}=\dfrac{3}{r(r+1)(r+2)(r+3)}, so P_3=\dfrac13\cdot\dfrac{1}{1\cdot2\cdot3}=\dfrac1{18}
M1A1
1=\dfrac{1}{1\times1!}, \dfrac14=\dfrac{1}{2\times2!}, \dfrac1{18}=\dfrac{1}{3\times3!}, so conjecture P_k=\dfrac{1}{k\cdot k!}
A1
(e) Let f(r)=\dfrac{1}{r(r+1)\cdots(r+k-1)}, a product of k consecutive integers
M1
f(r)-f(r+1)=\dfrac{(r+k)-r}{r(r+1)\cdots(r+k)}=\dfrac{k}{r(r+1)\cdots(r+k)}
A1
\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+1)\cdots(r+k)}=\frac1k\bigl(f(1)-f(n+1)\bigr), with f(1)=\dfrac{1}{k!}
M1
f(n+1)\to0 as n\to\infty, so P_k=\dfrac{1}{k}\cdot\dfrac{1}{k!}=\dfrac{1}{k\cdot k!}
A1

Final answer

  • (a) P_1=1
  • (c) \frac14-\frac{1}{2(n+1)(n+2)}, P_2=\frac14
  • (d) P_3=\frac1{18}, P_k=\frac{1}{k\cdot k!}
  • (e) f(r)=\frac{1}{r(r+1)\cdots(r+k-1)} gives f(r)-f(r+1)=\frac{k}{r(r+1)\cdots(r+k)}

G2 Small terms, endless sums

G2 Worked example

[9 marks]
For n\in\mathbb{Z}^+, let H_n=1+\dfrac12+\dfrac13+\dots+\dfrac1n.
(a) Explain why \dfrac15+\dfrac16+\dfrac17+\dfrac18>\dfrac12.
[1]
(b) For k\ge1, consider the 2^{k-1} terms from \dfrac{1}{2^{k-1}+1} to \dfrac{1}{2^k}. Show that they add to at least \dfrac12, and hence show that H_{2^m}\ge1+\dfrac m2 for every m\in\mathbb{Z}^+.
[3]
(c) By grouping the terms from 1 to \dfrac{1}{2^m-1} in a similar way, show that H_{2^m-1}\le m.
[2]
(d) Use (b) and (c) to give a range of values for the least n with H_n>5. Then find this n with your GDC.
[3]
Show solutionHide solution
(a) Each of the four terms is at least \frac18, and three of them are greater, so the total is greater than 4\times\frac18=\frac12
R1
(b) There are 2^k-2^{k-1}=2^{k-1} terms, and each is at least \dfrac1{2^k}, so their total is at least 2^{k-1}\times\dfrac{1}{2^k}=\dfrac12
M1A1
H_{2^m}=1+\dfrac12+\left(\dfrac13+\dfrac14\right)+\dots+\left(\dfrac{1}{2^{m-1}+1}+\dots+\dfrac1{2^m}\right): the 1 and then m blocks, k=1,\dots,m, each at least \frac12, so H_{2^m}\ge1+\dfrac m2
R1AG
(c) For j=0,1,\dots,m-1, the block from \dfrac1{2^j} to \dfrac{1}{2^{j+1}-1} has 2^j terms, each at most \dfrac1{2^j}, so it adds to at most 1
M1
These m blocks make up H_{2^m-1}, so H_{2^m-1}\le m
A1AG
(d) From (b) with m=8: H_{256}\ge5, and the inequality is strict because every block with k\ge2 has a term greater than \frac{1}{2^k}. So the least n is at most 256
A1
From (c) with m=5: H_{31}\le5, and H_n<H_{31} for n<31, so the least n is at least 32
A1
GDC: H_{82}=4.990\ldots, H_{83}=5.002\ldots, so the least n is 83
A1

Final answer

  • (d) 32\le n\le256
  • GDC: n=83 (H_{82}=4.990, H_{83}=5.002)

G2 Investigation

[12 marks]
This investigation looks at \displaystyle\sum_{r=1}^{\infty}\frac{1}{r^2}. You may assume that if a series of positive terms has partial sums that all stay below some fixed number, then the series has a sum to infinity, and that sum is not greater than that number.
(a) Show that, for every integer r\ge2, \dfrac1r-\dfrac1{r+1}<\dfrac1{r^2}<\dfrac{1}{r-1}-\dfrac1r.
[2]
(b) Hence show that \displaystyle\sum_{r=1}^{n}\frac1{r^2}<2-\frac1n for n\ge2. Explain why the series has a sum to infinity L, with L\le2.
[3]
(c) For m\in\mathbb{Z}^+, let A_m=\displaystyle\sum_{r=1}^{m}\frac1{r^2}. By using the inequalities in (a) only for the terms with r>m, show that A_m+\dfrac{1}{m+1}<L<A_m+\dfrac1m.
[3]
(d) Take m=10 to find an interval of width less than 0.01 that contains L. Give the endpoints to four decimal places.
[2]
(e) Find the least value of m for which the interval in (c) has width less than 10^{-4}.
[2]
Show solutionHide solution
(a) Over a common denominator, the left difference is \dfrac{1}{r(r+1)} and the right difference is \dfrac{1}{r(r-1)}
M1
For r\ge2 the three denominators are positive and r(r-1)<r\cdot r<r(r+1); a larger positive denominator gives a smaller fraction, which is the result
A1AG
(b) Keep the first term and use the right-hand inequality from r=2 on: 1+\frac14+\frac19+\dots+\frac1{n^2}<1+\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\dots+\left(\frac1{n-1}-\frac1n\right)
M1
The brackets telescope to 1-\dfrac1n, so the partial sum is less than 2-\dfrac1n
A1AG
The terms are positive and every partial sum is below 2, so by the given fact the series has a sum to infinity L\le2
R1
(c) L=A_m+\displaystyle\sum_{r=m+1}^{\infty}\frac1{r^2}. Upper: \displaystyle\sum_{r=m+1}^{\infty}\frac1{r^2}<\sum_{r=m+1}^{\infty}\left(\frac1{r-1}-\frac1r\right)=\frac1m
M1A1
Lower: \displaystyle\sum_{r=m+1}^{\infty}\frac1{r^2}>\sum_{r=m+1}^{\infty}\left(\frac1r-\frac1{r+1}\right)=\frac1{m+1}, so A_m+\dfrac1{m+1}<L<A_m+\dfrac1m
A1AG
Note: accept \le in place of < in the limit; the telescoped tails are \frac1m-\frac1N and \frac1{m+1}-\frac1{N+1} before N\to\infty.
(d) A_{10}=1.549767\ldots
M1
1.6406<L<1.6498 (endpoints 1.640677\ldots and 1.649767\ldots; width \frac1{110})
A1
(e) width =\dfrac1m-\dfrac1{m+1}=\dfrac{1}{m(m+1)}<10^{-4}\iff m(m+1)>10\,000
M1
99\times100=9900, 100\times101=10\,100, so m=100
A1

Final answer

  • (b) L\le2
  • (d) 1.6406<L<1.6498
  • (e) m=100

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 12.

JobTI-84 Plus CE keystrokes
Solve an equationMATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect
IntersectionsGraph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing

From the book

Photon Guidebook: Mathematics Analysis and Approaches SL for the IB Diploma

Chapter 1 teaches sequences and series from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book

Questions © 2026 Photon Academy IB Tuition. Original IB-style questions; not IB past-paper questions. This site is independent and not endorsed by the IB.