Free worked solutions · 2026 Edition

Chapter 1: Sequences and series worked solutions

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

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Basics

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

Question 1

[6 marks]

The nth term of an arithmetic sequence is u_n=3n-41.

(a) Write down the first term and the common difference.

[2]

(b) Find which term of the sequence is equal to 52.

[2]

(c) Find \displaystyle\sum_{n=1}^{31}(3n-41).

[2]

Show solutionHide solution

(a) u_1=-38

d=3

(b) 3n-41=52

n=31, the 31st term

(c) \dfrac{31}{2}(-38+52)

=217

Note: accept \dfrac{31}{2}\left(2(-38)+30(3)\right)=217.

Final answer

(a) u_1=-38, d=3
(b) the 31st term
(c) 217

Question 2

[6 marks]

Consider all the multiples of 9 that lie between 200 and 600.

(a) Write down the smallest and the largest of these multiples.

[2]

(b) Find how many of these multiples there are.

[2]

(c) Find the sum of these multiples.

[2]

Show solutionHide solution

(a) smallest 207

largest 594

(b) 207+9(n-1)=594

n=44

(c) \dfrac{44}{2}(207+594)

=17\,622

Note: in (b) accept 66-23+1=44.

Final answer

(a) 207 and 594
(b) 44
(c) 17\,622

Question 3

[5 marks]

The third term of an arithmetic sequence is 10, and the total of its first six terms is 69.

Find u_1 and d.

[5]

Show solutionHide solution

u_1+2d=10

\dfrac{6}{2}(2u_1+5d)=69

2u_1+5d=23

attempt to solve the two equations, e.g. 2(10-2d)+5d=23

d=3,\ u_1=4

Final answer

u_1=4, d=3

Question 4

[7 marks]

In a geometric sequence, u_3=50 and u_7=0.08.

(a) Find the two possible values of the common ratio.

[3]

(b) Show that both sequences have the same first term, and find it.

[2]

(c) Find u_6 for each sequence.

[2]

Show solutionHide solution

(a) \dfrac{u_7}{u_3}=r^{4}=\dfrac{0.08}{50}=0.0016

r^{4}=\dfrac{16}{10000}=\left(\dfrac{2}{10}\right)^{4}

r=0.2 or r=-0.2

(b) u_1r^{2}=50 and r^{2}=0.04 in both cases

u_1=\dfrac{50}{0.04}=1250

(c) u_6=1250\times(\pm 0.2)^{5}

u_6=0.4 (when r=0.2) and u_6=-0.4 (when r=-0.2)

Final answer

(a) r=\pm0.2
(b) u_1=1250
(c) u_6=0.4 or u_6=-0.4

Question 5

[6 marks]

The first term of a geometric sequence is 40 and the common ratio is 1.15.

(a) Show that the nth term is greater than 1000 when n-1>\dfrac{\ln 25}{\ln 1.15}.

[3]

(b) Hence find the first term of the sequence that is greater than 1000, and state its position in the sequence.

[3]

Show solutionHide solution

(a) 40\times 1.15^{n-1}>1000

1.15^{n-1}>25

(n-1)\ln 1.15>\ln 25 and \ln 1.15>0

n-1>\dfrac{\ln 25}{\ln 1.15}

(b) n-1>23.03\dots \Rightarrow n>24.03\dots (A1)

n=25 (check: u_{24}=995.7\dots, u_{25}=1145.0\dots)

the 25th term, u_{25}=1150 (1145.007\dots)

Note: Accept 1145, 1145.0 or 1150 for u_{25}.

Final answer

(b) u_{25}=1145.0\ldots\approx1150, the 25th term

Question 6

[5 marks]

An infinite geometric series has first term 10 and common ratio \dfrac{x+2}{5}.

(a) Find the set of values of x for which the series has a sum to infinity.

[3]

(b) Find the sum to infinity when x=1.

[2]

Show solutionHide solution

(a) \left|\dfrac{x+2}{5}\right|<1

-5<x+2<5 (A1)

-7<x<3

(b) r=\dfrac{3}{5}, \ S_\infty=\dfrac{10}{1-\frac{3}{5}}

=25

Final answer

(a) -7<x<3
(b) 25

Question 7

[5 marks]

Consider the number 2.1\dot{3}=2.13333\dots

(a) Write 0.03333\dots as an infinite geometric series and state its first term and common ratio.

[2]

(b) Hence express 2.1\dot{3} as a fraction in its lowest terms.

[3]

Show solutionHide solution

(a) 0.03+0.003+0.0003+\dots

u_1=0.03, \ r=0.1

(b) 0.0333\dots=\dfrac{0.03}{1-0.1}=\dfrac{1}{30} M1A1

2.1\dot{3}=\dfrac{21}{10}+\dfrac{1}{30}=\dfrac{64}{30}=\dfrac{32}{15}

Final answer

(a) 0.03+0.003+0.0003+\cdots, first term 0.03, ratio 0.1
(b) \frac{32}{15}

Question 8

[8 marks]

The sum of the first n terms of an arithmetic sequence is S_n=7n-n^2.

(a) Find S_1 and S_2, and hence write down u_1 and u_2.

[3]

(b) By simplifying S_n-S_{n-1}, find an expression for u_n in terms of n.

[3]

(c) Find the largest value of S_n, and the values of n for which it occurs. Justify your answer.

[2]

Show solutionHide solution

(a) S_1=6,\ S_2=10

u_1=6

u_2=10-6=4

(b) S_n-S_{n-1}=7n-n^2-\left(7(n-1)-(n-1)^2\right)

=7-(2n-1)

u_n=8-2n

(c) u_n>0 for n\le 3, u_4=0 and u_n<0 for n\ge 5, so the sum stops increasing after the third term

largest value S_3=S_4=12, when n=3 or n=4

Final answer

(a) S_1=6, S_2=10; u_1=6, u_2=4
(b) u_n=8-2n
(c) 12, when n=3 or n=4

Tutorial

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

Question 9

[5 marks]

For an arithmetic sequence, u_4 + u_9 = 41 and S_{10} = 175.

Find the first term and the common difference.

Show solutionHide solution

u_4 + u_9 = (u_1 + 3d) + (u_1 + 8d) = 2u_1 + 11d = 41.

[2]

S_{10} = 5(2u_1 + 9d) = 175 \Rightarrow 2u_1 + 9d = 35.

[1]

Subtracting: 2d = 6, so d = 3 and u_1 = 4.

[2]

Final answer

u_1=4, d=3

Question 10

[5 marks]

The numbers x, x+6 and 4x, where x \neq 0, are three consecutive terms of a geometric sequence.

(a) Find the possible values of x.

[3]

(b) For each value of x, write down the common ratio.

[2]

Show solutionHide solution

(a) Middle term squared equals the product of its neighbours: (x + 6)^2 = x \cdot 4x.

[1]

x^2 + 12x + 36 = 4x^2 \Rightarrow 3x^2 - 12x - 36 = 0 \Rightarrow x^2 - 4x - 12 = 0 \Rightarrow (x - 6)(x + 2) = 0.

[1]

x = 6 or x = -2.

[1]

(b) x = 6: terms 6, 12, 24, so r = 2.

[1]

x = -2: terms -2, 4, -8, so r = -2.

[1]

Final answer

(a) x=6 or x=-2
(b) r=2 for x=6; r=-2 for x=-2

Question 11

[7 marks]

An arithmetic sequence has first term 33 and common difference -2.

(a) Show that S_n = n(34 - n).

[2]

(b) Find the values of n for which S_n = 280.

[2]

(c) Explain, by considering the terms of the sequence, why there are two such values of n.

[1]

(d) Find the maximum value of S_n, and the value of n at which it occurs.

[2]

Show solutionHide solution

(a) S_n = \dfrac{n}{2}\bigl(66 - 2(n - 1)\bigr)

[1]

= \dfrac{n}{2}(68 - 2n) = n(34 - n).

[1]

(b) 34n - n^2 = 280 \Rightarrow n^2 - 34n + 280 = 0 \Rightarrow (n - 14)(n - 20) = 0

[1]

n = 14 or n = 20.

[1]

(c) u_{15}, \ldots, u_{20} are 5, 3, 1, -1, -3, -5, which add to 0, so S_{20} = S_{14}.

[1]

(d) u_n = 35 - 2n > 0 \iff n \le 17, so S_n is greatest at the last positive term, n = 17.

[1]

S_{17} = 17 \times 17 = 289.

[1]

Final answer

(b) n=14 or n=20
(c) u_{15} to u_{20} add to 0
(d) 289, at n=17

Question 12

[6 marks]

An infinite geometric series has first term 12 and sum to infinity 20.

(a) Find the common ratio.

[2]

(b) Find the third term.

[1]

(c) A new series is formed by squaring each term of the original series. Find the sum to infinity of the new series.

[3]

Show solutionHide solution

(a) \dfrac{12}{1 - r} = 20

[1]

1 - r = \frac35 \Rightarrow r = \frac25.

[1]

(b) u_3 = 12\left(\frac25\right)^2 = \dfrac{48}{25}.

[1]

(c) The squares form a GP with first term 144 and ratio r^2 = \frac{4}{25} (and |r^2| < 1).

[2]

S_\infty = \dfrac{144}{21/25} = \dfrac{1200}{7}.

[1]

Final answer

(a) \frac25
(b) \frac{48}{25}
(c) \frac{1200}{7}

Question 13

[6 marks]

A geometric series has first term 40 and common ratio \frac12.

(a) Show that S_\infty - S_n = 80\left(\dfrac12\right)^n.

[3]

(b) Find the least value of n for which S_n > 79.95.

[3]

Show solutionHide solution

(a) S_\infty = \dfrac{40}{1 - \frac12} = 80.

[1]

S_n = \dfrac{40\bigl(1 - (\frac12)^n\bigr)}{\frac12} = 80 - 80\left(\frac12\right)^n.

[2]

So S_\infty - S_n = 80\left(\dfrac12\right)^n.

(b) S_n > 79.95 \iff 80\left(\frac12\right)^n < 0.05 \iff 2^n > 1600.

[2]

2^{10} = 1024 < 1600 < 2048 = 2^{11}, so n = 11.

[1]

Final answer

(b) n=11

Question 14

[8 marks]

Consider the infinite geometric series

1 + \frac{x+1}{x-3} + \left(\frac{x+1}{x-3}\right)^2 + \cdots, \qquad x \neq 3.

(a) Show that the series converges if and only if x < 1.

[4]

(b) Show that, for x < 1, the sum to infinity is \dfrac{3-x}{4}.

[2]

(c) Explain why the sum to infinity can never equal \frac12.

[2]

Show solutionHide solution

(a) Converges \iff \left|\dfrac{x+1}{x-3}\right| < 1 \iff |x + 1| < |x - 3| (as x \neq 3).

[1]

Both sides are non-negative, so square: (x+1)^2 < (x-3)^2

[1]

x^2 + 2x + 1 < x^2 - 6x + 9 \Rightarrow 8x < 8

[1]

\iff x < 1. Every step is reversible, so the series converges if and only if x < 1.

[1]

(b) S_\infty = \dfrac{1}{1 - \frac{x+1}{x-3}} = \dfrac{x - 3}{(x - 3) - (x + 1)}

[1]

= \dfrac{x-3}{-4} = \dfrac{3-x}{4}.

[1]

(c) \dfrac{3-x}{4} = \dfrac12 \Rightarrow x = 1.

[1]

But the series converges only for x < 1, so this value is not allowed (for x < 1, S_\infty > \frac12).

[1]

Final answer

(b) \frac{3-x}{4}
(c) it would need x=1, which is excluded

Question 15

[7 marks]

The sum of the first n terms of a sequence is S_n = 3\times 5^{n} + k, where k is a constant.

(a) Find u_1 in terms of k.

[1]

(b) Show that u_n = 12\times 5^{n-1} for n \ge 2.

[2]

(c) Find the value of k for which the sequence is geometric, and write down the common ratio in this case.

[2]

(d) For this value of k, find the least value of n for which S_n > 10\,000.

[2]

Show solutionHide solution

(a) u_1 = S_1 = 15 + k.

[1]

(b) For n \ge 2: u_n = S_n - S_{n-1} = (3 \times 5^{n} + k) - (3 \times 5^{n-1} + k) = 3 \times 5^{n-1}(5 - 1)

[1]

= 12 \times 5^{n-1} (the constant k cancels).

[1]

(c) From u_2 onwards each term is 5 times the previous one, so the sequence is geometric exactly when u_2 = 5u_1 as well: 60 = 5(15 + k).

[1]

k = -3, and the common ratio is 5.

[1]

(d) 3 \times 5^{n} - 3 > 10\,000 \Rightarrow 5^{n} > 3334.\overline{3}

[1]

5^5 = 3125, 5^6 = 15\,625, so n = 6.

[1]

Final answer

(a) 15+k
(c) k=-3, r=5
(d) n=6

Question 16

[8 marks]

A geometric sequence has first term 8 and common ratio r, where r \neq 1. The 1st, 3rd and 4th terms of this geometric sequence are three consecutive terms of an arithmetic sequence.

(a) Show that r^3 - 2r^2 + 1 = 0.

[2]

(b) Show that r^3 - 2r^2 + 1 = (r-1)(r^2 - r - 1), and hence find the exact possible values of r.

[3]

(c) Only one of these values of r gives a geometric series with a sum to infinity. Identify this value, and show that the sum to infinity is 4\left(\sqrt5 - 1\right).

[3]

Show solutionHide solution

(a) Terms 8, 8r^2, 8r^3 in arithmetic progression: 2(8r^2) = 8 + 8r^3

[2]

\Rightarrow r^3 - 2r^2 + 1 = 0.

(b) (r - 1)(r^2 - r - 1) = r^3 - r^2 - r - r^2 + r + 1 = r^3 - 2r^2 + 1.

[1]

r \neq 1, so r^2 - r - 1 = 0

[1]

r = \dfrac{1 \pm \sqrt5}{2}.

[1]

(c) Since 2 < \sqrt5 < 3: \frac{1+\sqrt5}{2} > \frac32 > 1, while -1 < \frac{1-\sqrt5}{2} < -\frac12, so r = \dfrac{1 - \sqrt5}{2}.

[1]

S_\infty = \dfrac{8}{1 - \frac{1-\sqrt5}{2}} = \dfrac{16}{1 + \sqrt5}

[1]

= \dfrac{16(\sqrt5 - 1)}{(\sqrt5 + 1)(\sqrt5 - 1)} = \dfrac{16(\sqrt5 - 1)}{4} = 4(\sqrt5 - 1).

[1]

Final answer

(b) r=\frac{1\pm\sqrt5}{2}
(c) r=\frac{1-\sqrt5}{2}

Question 17

[6 marks]

A geometric sequence u_n has positive terms, with u_1 = 12 and u_5 = \frac34.

(a) Find the common ratio.

[2]

The sequence v_n is defined by v_n = \log_2 u_n.

(b) Show that v_n is an arithmetic sequence with common difference -1.

[2]

(c) Find \displaystyle\sum_{n=1}^{12} v_n, giving your answer in the form a + b\log_2 3, where a, b \in \mathbb{Z}.

[2]

Show solutionHide solution

(a) r^4 = \dfrac{3/4}{12} = \dfrac{1}{16}

[1]

terms positive, so r = \frac12.

[1]

(b) v_{n+1} - v_n = \log_2 u_{n+1} - \log_2 u_n = \log_2\dfrac{u_{n+1}}{u_n}

[1]

= \log_2 \frac12 = -1, a constant, so v_n is arithmetic with d = -1.

[1]

(c) v_1 = \log_2 12 = 2 + \log_2 3; \displaystyle\sum_{n=1}^{12} v_n = \frac{12}{2}\bigl(2(2 + \log_2 3) + 11(-1)\bigr)

[1]

= 6(-7 + 2\log_2 3) = -42 + 12\log_2 3.

[1]

Final answer

(a) \frac12
(c) -42+12\log_2 3

Question 18

[6 marks]

A theatre has 25 rows of seats. Row 1 has 18 seats, and each row after the first has 2 more seats than the row in front of it. A ticket in row 1 costs 60 dollars, and the ticket price in each row after the first is 6\% lower than in the row in front of it.

(a) Find the total number of seats in the theatre.

[2]

(b) Write down an expression, in terms of r, for the income from row r when all of its seats are sold. Hence find the total income when every seat in the theatre is sold.

[2]

(c) Find which row gives the greatest income when all of its seats are sold.

[2]

Show solutionHide solution

Seats in row r: 18 + 2(r - 1) = 2r + 16 (arithmetic). Price in row r: 60(0.94)^{r-1} (geometric).

(a) S_{25} = \dfrac{25}{2}\bigl(2(18) + 24(2)\bigr)

[1]

= 1050 seats.

[1]

(b) Income from row r: (2r + 16) \times 60(0.94)^{r-1} dollars. GDC: \displaystyle\sum_{r=1}^{25} 60(2r + 16)(0.94)^{r-1}

[1]

= 28\,184.259\ldots, so $28\,184.26.

[1]

(c) GDC table of 60(2r + 16)(0.94)^{r-1}: row 7: $1241.77; row 8: $1245.08; row 9: $1243.52

[1]

(the seat count grows by a shrinking percentage while the price falls by a steady 6\%, so the income rises and then falls) the greatest income is from row 8.

[1]

Final answer

(a) 1050
(b) 28\,184.26 dollars
(c) row 8

Mastery

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

Question 19

[7 marks]

For an arithmetic sequence, the ratio of the sum of the first eight terms to the sum of the first four terms is 7 : 3. The third term is 25.

(a) Find the first term and the common difference.

[5]

(b) Find the sum of the first fifteen terms.

[2]

Show solutionHide solution

(a) S_8 = 4(2u_1 + 7d), \ S_4 = 2(2u_1 + 3d)

\dfrac{S_8}{S_4} = \dfrac{7}{3} \implies 3\cdot 4(2u_1 + 7d) = 7\cdot 2(2u_1 + 3d)

24u_1 + 84d = 28u_1 + 42d \implies 2u_1 = 21d

u_3 = u_1 + 2d = 25: \dfrac{21d}{2} + 2d = 25 \implies d = 2

u_1 = 21,\qquad d = 2

A1[5]

(b) S_{15} = \dfrac{15}{2}\left(42 + 14(2)\right) = 525

M1A1[2]

Final answer

(a) u_1=21, d=2
(b) 525

Question 20

[9 marks]

An infinite geometric series has first term a, common ratio r, where -1 < r < 1, r \neq 0, and sum to infinity S.

(a) Show that the sum of the terms in the even positions, u_2 + u_4 + u_6 + \cdots, is equal to \dfrac{r}{1+r}\,S.

[3]

(b) Given that S = 12 and that the sum to infinity of the squares of the terms is 48, find the value of a and the value of r.

[4]

(c) For the series in part (b), find the sum to infinity of the cubes of the terms.

[2]

Show solutionHide solution

(a) u_2 + u_4 + \cdots = ar + ar^3 + \cdots, a GP with first term ar and ratio r^2 (|r^2| < 1).

[1]

Sum = \dfrac{ar}{1 - r^2} = \dfrac{a}{1 - r} \cdot \dfrac{r}{1 + r}

[1]

= \dfrac{r}{1 + r}\,S.

[1]

(b) \dfrac{a}{1 - r} = 12 and \dfrac{a^2}{1 - r^2} = 48

[1]

Dividing the second by the first: \dfrac{a}{1 + r} = 4.

[1]

a = 12(1 - r) = 4(1 + r) \Rightarrow 16r = 8, so r = \dfrac12

[1]

and a = 6.

[1]

(c) Cubes: first term 216, ratio \frac{1}{8}: \dfrac{216}{1 - \frac{1}{8}}

[1]

= \dfrac{1728}{7}.

[1]

Final answer

(b) a=6, r=\frac12
(c) \frac{1728}{7}

Question 21

[14 marks]

A patient takes a 200\,\text{mg} tablet at 8 am every day. Between one tablet and the next, 30\% of the drug in the body is removed. Let A_n be the amount of drug (in mg) in the body immediately after the nth tablet is taken.

(a) Find A_1, A_2 and A_3.

[3]

(b) Show that A_n = \dfrac{2000}{3}\left(1 - 0.7^{\,n}\right).

[3]

(c) Find the value that A_n approaches in the long term.

[2]

The dose is changed to x\,\text{mg} per day so that the long-term value of A_n becomes 800\,\text{mg}.

(d) Find x.

[2]

(e) Returning to the 200\,\text{mg} dose, find the least value of n for which A_n exceeds 95\% of its long-term value.

[4]

Show solutionHide solution

(a) A_1 = 200

A_2 = 0.7(200) + 200 = 340

A_3 = 0.7(340) + 200 = 438

A1[3]

(b) A_n = 200 + 200(0.7) + \cdots + 200(0.7)^{n-1}

M1A1

This is geometric with a = 200, r = 0.7: A_n = \dfrac{200\left(1 - 0.7^{n}\right)}{0.3} = \dfrac{2000}{3}\left(1 - 0.7^{n}\right)

A1AG[3]

(c) As n \to \infty, 0.7^{n} \to 0, so A_n \to \dfrac{2000}{3} \approx 667\,\text{mg}

M1A1[2]

(d) Long-term value = \dfrac{x}{0.3} = 800

x = 240\,\text{mg}

A1[2]

(e) \dfrac{2000}{3}\left(1 - 0.7^{n}\right) > 0.95 \times \dfrac{2000}{3} \implies 0.7^{n} < 0.05

M1A1

n > \dfrac{\ln 0.05}{\ln 0.7} = 8.40\ldots

n = 9

A1[4]

Final answer

(a) 200, 340, 438
(c) \frac{2000}{3}\approx667 mg
(d) 240 mg
(e) n=9

Question 22

[13 marks]

A designer draws a fractal tree. Stage 0 is a vertical trunk of length 1.5 m. At each stage k \ge 1, every branch drawn at the previous stage splits into two new branches, and each new branch is 0.45 times as long as the branch it grows from.

(a) Show that the total length of the branches drawn at stage k is 1.5(0.9)^k m.

[2]

(b) Find the total length of the tree after stage 10 has been drawn.

[2]

(c) Show that the total length of the tree can never exceed 15 m.

[2]

(d) Find the least value of n for which the total length after stage n exceeds 14 m.

[3]

In a second design each new branch is p times as long as the branch it grows from, where 0 < p < 1.

(e) (i) Find the set of values of p for which the total length of the tree remains finite.

[2]

(ii) Find the value of p for which the total length of the tree tends to 30 m.

[2]

Show solutionHide solution

(a) At stage k there are 2^k new branches

[1]

each of length 1.5(0.45)^k, so total 2^k \times 1.5(0.45)^k = 1.5(0.9)^k m.

[1]

(b) Stages 0 to 10: \displaystyle\sum_{k=0}^{10} 1.5(0.9)^k = \frac{1.5(1 - 0.9^{11})}{0.1}

[1]

= 10.292\ldots \approx 10.3 m.

[1]

(c) The stage totals form a GP with |r| = 0.9 < 1, so the total length tends to S_\infty = \dfrac{1.5}{1 - 0.9} = 15 m

[1]

and every partial sum 15(1 - 0.9^{n+1}) is less than 15.

[1]

(d) 15(1 - 0.9^{n+1}) > 14 \Rightarrow 0.9^{n+1} < \dfrac{1}{15}

[1]

n + 1 > \dfrac{\ln(1/15)}{\ln 0.9} = 25.7\ldots (after stage 24: 13.92 m; after stage 25: 14.03 m)

[1]

n = 25.

[1]

(e) (i) Stage totals 1.5(2p)^k: a GP with ratio 2p; finite total requires 2p < 1, so 0 < p < \frac12.

[2]

(ii) \dfrac{1.5}{1 - 2p} = 30 \Rightarrow 1 - 2p = 0.05

[1]

p = 0.475.

[1]

Final answer

(b) 10.3 m
(d) n=25
(e)(i) 0<p<\frac12
(ii) p=0.475

Question 23

[16 marks]

Logs are stacked in rows. The bottom row has 25 logs, and each row has 2 fewer logs than the row below it, until the top row has 1 log. Rows are numbered r = 1, 2, 3, \ldots from the bottom.

(a) Write down the number of logs in row 3.

[1]

(b) Show that row r contains 27 - 2r logs.

[2]

(c) Find the number of rows in the complete stack.

[2]

(d) Show that the number of logs in the bottom n rows is n(26 - n).

[3]

(e) Find the total number of logs in the complete stack.

[2]

A second stack is built the same way but its bottom row has 31 logs and each row has 3 fewer than the row below, ending with a top row of 1 log.

(f) (i) Find the number of rows in the second stack.

[2]

(ii) Find the total number of logs in the second stack.

[2]

(iii) Each log has mass 18\,\text{kg}. Find the difference in mass between the two stacks.

[2]

Show solutionHide solution

(a) 25 - 2 - 2 = 21

A1[1]

(b) Arithmetic with u_1 = 25, d = -2: u_r = 25 - 2(r-1)

= 25 - 2r + 2

u_r = 27 - 2r \qquad \textbf{AG}

[2]

(c) 27 - 2r = 1 \implies r = 13

M1A1[2]

(d) S_n = \dfrac{n}{2}\left(25 + (27 - 2n)\right)

M1A1

= \dfrac{n}{2}(52 - 2n)

= n(26 - n) \qquad \textbf{AG}

[3]

(e) S_{13} = 13(13) = 169

M1A1[2]

(f)(i) 31 - 3(r-1) = 1 \implies r = 11 rows

M1A1

(ii) S_{11} = \dfrac{11}{2}(31 + 1) = 176 logs

M1A1

(iii) Difference = 176 - 169 = 7 logs, so 7 \times 18 = 126\,\text{kg}

M1A1[6]

Final answer

(a) 21
(c) 13
(e) 169
(f) 11 rows; 176 logs; 126 kg

Question 24

[16 marks]

A geometric sequence has g_n = 4 \times 3^{\,n-1} for n \in \mathbb{Z}^{+}. Let w_n = \ln g_n.

(a) Show that \{w_n\} is an arithmetic sequence, stating its first term and common difference.

[3]

(b) Find \displaystyle\sum_{n=1}^{20} w_n in the form p\ln 2 + q\ln 3, where p, q \in \mathbb{Z}^{+}.

[4]

(c) Find the smallest value of n for which g_n > 10^{6}.

[3]

Let P_n = g_1 g_2 g_3 \cdots g_n.

(d) Show that \ln P_n = n\ln 4 + \dfrac{n(n-1)}{2}\ln 3.

[3]

(e) Hence find the least value of n for which P_n > 10^{30}.

[3]

Show solutionHide solution

(a) w_n = \ln 4 + (n-1)\ln 3

M1A1

w_{n+1} - w_n = \ln 3 is constant, so the sequence is arithmetic with w_1 = \ln 4 and d = \ln 3.

A1[3]

(b) S_{20} = \dfrac{20}{2}\left(2\ln 4 + 19\ln 3\right)

M1A1

= 20\ln 4 + 190\ln 3 = 40\ln 2 + 190\ln 3

M1A1[4]

(c) 4\cdot 3^{n-1} > 10^{6} \implies n - 1 > \dfrac{\ln(250\,000)}{\ln 3} = 11.3\ldots

M1A1

n = 13

A1[3]

(d) \ln P_n = \ln g_1 + \cdots + \ln g_n = \sum_{i=1}^{n} w_i

= \dfrac{n}{2}\left(2\ln 4 + (n-1)\ln 3\right)

= n\ln 4 + \dfrac{n(n-1)}{2}\ln 3

A1AG[3]

(e) Need n\ln 4 + \dfrac{n(n-1)}{2}\ln 3 > 30\ln 10 = 69.08\ldots

n = 10 gives 63.3\ldots; n = 11 gives 75.7\ldots

n = 11

A1[3]

Final answer

(a) w_1=\ln4, d=\ln3
(b) 40\ln2+190\ln3
(c) 13
(e) 11

Question 25

[14 marks]

The sum of the first n terms of a sequence u_n is given by S_n = 2n^2 - 11n, for n \in \mathbb{Z}^+.

(a) Find u_1 and u_2.

[2]

(b) Show that the sequence is arithmetic, and state its common difference.

[3]

(c) Given that S_k = 90, find the value of k.

[3]

(d) Find the value of \displaystyle\sum_{n=1}^{10} \left|u_n\right|.

[3]

(e) Show that \displaystyle\sum_{n=1}^{2m} (-1)^n u_n = 4m for all m \in \mathbb{Z}^+.

[3]

Show solutionHide solution

(a) u_1 = S_1 = 2 - 11 = -9.

[1]

u_2 = S_2 - S_1 = (8 - 22) - (-9) = -5.

[1]

(b) u_n = S_n - S_{n-1} = 2n^2 - 11n - \bigl(2(n-1)^2 - 11(n-1)\bigr)

[1]

= 4n - 13 (true also at n = 1).

[1]

u_{n+1} - u_n = 4, a constant, so arithmetic with d = 4.

[1]

(c) 2k^2 - 11k = 90 \Rightarrow 2k^2 - 11k - 90 = 0

[1]

(2k + 9)(k - 10) = 0

[1]

k \in \mathbb{Z}^+, so k = 10.

[1]

(d) u_1, u_2, u_3 = -9, -5, -1 are negative; u_4, \ldots, u_{10} are positive.

[1]

\displaystyle\sum_{n=1}^{10}|u_n| = (S_{10} - S_3) - S_3 = 90 - 2(-15)

[1]

= 120.

[1]

(e) Pair the terms: (-u_1 + u_2) + (-u_3 + u_4) + \cdots + (-u_{2m-1} + u_{2m})

[1]

Each bracket is u_{2j} - u_{2j-1} = d = 4

[1]

and there are m brackets, so the sum is 4m.

[1]

Final answer

(a) -9, -5
(b) d=4
(c) k=10
(d) 120
(e) proved in the solutions

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) r=1: 1-\frac13; r=2: \frac12-\frac14; r=3: \frac13-\frac15; r=4: \frac14-\frac16; …; r=n-1: \frac1{n-1}-\frac1{n+1}; r=n: \frac1n-\frac1{n+2}

Each negative term reappears as a positive term two rows later, so only 1+\frac12-\frac1{n+1}-\frac1{n+2} survives.

By (a), each term of the required sum is half a row, so halve the survivors

\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

(d) \dfrac34-S_n=\dfrac12\left(\dfrac1{n+1}+\dfrac1{n+2}\right)<0.01\iff\dfrac1{n+1}+\dfrac1{n+2}<0.02

n=98: \frac1{99}+\frac1{100}=0.02010\ldots>0.02; ; n=99: \frac1{100}+\frac1{101}=0.01990\ldots<0.02

n=99

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

\displaystyle\sum_{r=1}^{\infty}\frac1{r(r+3)}=\frac13\left(1+\frac12+\frac13\right)=\frac{11}{18}

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}

\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)}

=\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)

\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)}

\dfrac{1}{2(n+1)(n+2)}\to0, so P_2=\dfrac14

(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!}

(e) Let f(r)=\dfrac{1}{r(r+1)\cdots(r+k-1)}, a product of k consecutive integers

f(r)-f(r+1)=\dfrac{(r+k)-r}{r(r+1)\cdots(r+k)}=\dfrac{k}{r(r+1)\cdots(r+k)}

\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!}

f(n+1)\to0 as n\to\infty, so P_k=\dfrac{1}{k}\cdot\dfrac{1}{k!}=\dfrac{1}{k\cdot k!}

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

(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

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

From (c) with m=5: H_{31}\le5, and H_n<H_{31} for n<31, so the least n is at least 32

GDC: H_{82}=4.990\ldots, H_{83}=5.002\ldots, so the least n is 83

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)}

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)

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

(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

1.6406<L<1.6498 (endpoints 1.640677\ldots and 1.649767\ldots; width \frac1{110})

(e) width =\dfrac1m-\dfrac1{m+1}=\dfrac{1}{m(m+1)}<10^{-4}\iff m(m+1)>10\,000

99\times100=9900, 100\times101=10\,100, so m=100

Remark: Euler showed that L=\dfrac{\pi^2}{6}=1.644934\ldots, which lies in the interval from (d).

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 19.

Job	TI-84 Plus CE keystrokes
Solve an equation	`MATH, 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`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

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