S5 Mathematics Practice Book · Unit 2: Sequences

Unit 2 Sequences

Key unit competence: Use arithmetic, geometric and harmonic sequences, including their convergence, in solving related problems.

Key ideas

A sequence $(u_{n})$ is a function from $N$ to $R$ . It can be given by a general term $u_{n} = f (n)$ or by recurrence $u_{n + 1} = g (u_{n})$ .
Monotone: increasing if $u_{n + 1} \geq u_{n}$ for all $n$ (study the sign of $u_{n + 1} - u_{n}$ , or $\frac{u _{n + 1}}{u _{n}}$ for positive terms); decreasing if $u_{n + 1} \leq u_{n}$ .
Arithmetic sequence: $u_{n + 1} = u_{n} + d$ ; $u_{n} = u_{1} + (n - 1) d$ ; $S_{n} = \frac{n}{2} (u_{1} + u_{n}) = \frac{n}{2} [2 u_{1} + (n - 1) d]$ . Three numbers $a, b, c$ are in AP if $2 b = a + c$ (arithmetic mean).
Geometric sequence: $u_{n + 1} = r u_{n}$ ; $u_{n} = u_{1} r^{n - 1}$ ; $S_{n} = u_{1} \frac{1 - r ^{n}}{1 - r}$ ( $r \neq = 1$ ). $a, b, c$ in GP if $b^{2} = a c$ (geometric mean $a c$ ).
Harmonic sequence: the reciprocals form an arithmetic sequence. The harmonic mean of $a$ and $b$ is $\frac{2 ab}{a + b}$ .
Convergence: $(u_{n})$ converges to $L$ if $u_{n} \to L$ . A geometric sequence with $∣ r ∣ < 1$ tends to 0; the infinite sum is $S_{\infty} = \frac{u _{1}}{1 - r}$ for $∣ r ∣ < 1$ .

Formulas to know

u_{n} = u_{1} + (n - 1) d, S_{n} = \frac{n}{2} (u_{1} + u_{n})

u_{n} = u_{1} r^{n - 1}, S_{n} = u_{1} \frac{1 - r ^{n}}{1 - r}

S_{\infty} = \frac{u _{1}}{1 - r} (∣ r ∣ < 1)

A M = \frac{a + b}{2}, GM = ab, H M = \frac{2 ab}{a + b}

Common misconceptions

"In an AP, $u_{n} = u_{1} + n d$ ." No: $u_{n} = u_{1} + (n - 1) d$ .
" $1 + 2 + 4 + 8 + \dots$ has a finite sum." No: $r = 2$ , so it diverges; $S_{\infty}$ needs $∣ r ∣ < 1$ .

Worked examples

Example 1. In an AP $u_{1} = 5$ and $d = 3$ . Find $u_{20}$ and $S_{20}$ .

Solution

$u_{20} = 5 + 19 \times 3 = 62$ ; $S_{20} = \frac{20}{2} (5 + 62) = 670$ .

Example 2. Find the sum to infinity of $8 + 4 + 2 + 1 + \dots$

Solution

$r = \frac{1}{2}$ : $S_{\infty} = \frac{8}{1 - \frac{1}{2}} = 16$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

Find the 15th term and the sum of the first 15 terms of $20, 17, 14, \dots$
Show answer
$d = - 3$ ; $u_{15} = 20 - 42 = - 22$ ; $S_{15} = \frac{15}{2} (20 - 22) = - 15$ .
Common mistake: Using $+ 3$ for the difference.
Prove that the sequence $u_{n} = \frac{2 n + 1}{n + 3}$ converges and give its limit.
Show answer
Divide by $n$ : $\frac{2 + \frac{1}{n}}{1 + \frac{3}{n}} \to 2$ .
Common mistake: Taking the ratio of the constant terms $\frac{1}{3}$ .
Find the sum of the first 8 terms of the GP $5, 10, 20, \dots$
Show answer
$5 \times \frac{2 ^{8} - 1}{2 - 1} = 1275$ .
Common mistake: Using $2^{7}$ instead of $2^{8}$ .
A population of 10 000 decreases by 10% per year. Write $u_{n}$ and find when it falls below 5 000.
Show answer
$u_{n} = 10000 \times 0. 9^{n}$ ; $0. 9^{n} < 0.5 \Rightarrow n > \frac{lo g 0.5}{lo g 0.9} \approx 6.6$ , so after 7 years.
Common mistake: Using $10000 - 1000 n$ (linear).
Find the sum to infinity of $27 - 9 + 3 - 1 + \dots$
Show answer
$r = - \frac{1}{3}$ ; $S_{\infty} = \frac{27}{1 + \frac{1}{3}} = \frac{81}{4} = 20.25$ .
Common mistake: Ignoring the negative ratio.
The 2nd and 5th terms of a harmonic sequence are $\frac{1}{4}$ and $\frac{1}{10}$ . Find the 8th term.
Show answer
Reciprocals form an AP: $4$ and $10$ , so $3 d = 6$ , $d = 2$ ; 8th reciprocal $= 4 + 6 \times 2 = 16$ . 8th term $\frac{1}{16}$ .
Common mistake: Applying the AP formula to the terms themselves.
Show that $u_{n} = 3 n - 2$ is an AP and find $S_{n}$ .
Show answer
$u_{n + 1} - u_{n} = 3$ (constant). $S_{n} = \frac{n}{2} (1 + 3 n - 2) = \frac{n ( 3 n - 1 )}{2}$ .
Common mistake: Checking only two terms.
Insert two geometric means between 3 and 81.
Show answer
$81 = 3 r^{3}$ , $r = 3$ : 9 and 27.
Common mistake: Inserting arithmetic means instead.

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Find the sum of the first 30 terms of $2, 5, 8, \dots$	a. $r = 2$ , $u_{1} = 3$
2. Find $x > 0$ so that $x$ , 6, 18 are consecutive terms of a geometric sequence.	b. $10$
3. A geometric sequence has $u_{3} = 12$ and $u_{6} = 96$ . Find $r$ and $u_{1}$ .	c. $1365$
4. For $1 + 2 + 4 + \dots$ , find the smallest $n$ with $S_{n} > 1000$ .	d. $x = 2$

Show answer

1 → c ( $1365$ ) · 2 → d ( $x = 2$ ) · 3 → a ( $r = 2$ , $u_{1} = 3$ ) · 4 → b ( $10$ )

Step 1. $u_{1} = 2$ , $d = 3$ , $u_{30} = 2 + 29 \times 3 = 89$ .
Step 2. $S_{30} = \frac{30}{2} (2 + 89) = 15 \times 91 = 1365$ .
Step 1. Geometric mean: $6^{2} = 18 x$ .
Step 2. $x = \frac{36}{18} = 2$ .
Step 1. $\frac{u _{6}}{u _{3}} = r^{3} = 8$ , so $r = 2$ .
Step 2. $u_{1} = \frac{u _{3}}{r ^{2}} = \frac{12}{4} = 3$ .
Step 1. $S_{n} = 2^{n} - 1$ .
Step 2. $2^{9} - 1 = 511$ , $2^{10} - 1 = 1023$ : $n = 10$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
10 and 64 · 10 and 8 · $4$ · $u_{n} = 4 n - 6$
1. Find the harmonic mean of 3 and 6. →
2. Find the arithmetic and geometric means of 4 and 16. →
3. $u_{1} = - 2$ and $u_{n + 1} = u_{n} + 4$ . Find the general term. →
Show answer
1: $4$ · 2: 10 and 8 · 3: $u_{n} = 4 n - 6$
- Step 1. $H = \frac{2 ab}{a + b}$ .
  Step 2. $\frac{2 \times 18}{9} = 4$ .
- Step 1. Arithmetic mean $\frac{4 + 16}{2} = 10$ .
  Step 2. Geometric mean $4 \times 16 = 8$ .
- Step 1. Arithmetic with $d = 4$ : $u_{n} = - 2 + 4 (n - 1)$ .
  Step 2. $= 4 n - 6$ .

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Find $lim_{n \to \infty} \frac{2 n + 1}{n + 3}$ .	a. $1683$
2. A population of 5 000 grows by 4% per year. Find it after 10 years.	b. $\approx 7401$
3. A student saves 1 000 FRW in month 1 and 200 FRW more each month. Find the total saved in 12 months.	c. $25200$ FRW
4. Find the sum of the multiples of 3 from 3 to 99.	d. $2$

Show answer

1 → d ( $2$ ) · 2 → b ( $\approx 7401$ ) · 3 → c ( $25200$ FRW) · 4 → a ( $1683$ )

Step 1. Divide by $n$ : $\frac{2 + \frac{1}{n}}{1 + \frac{3}{n}}$ .
Step 2. As $n \to \infty$ : $\frac{2}{1} = 2$ .
Step 1. Geometric: $u = 5000 \times 1.0 4^{10}$ .
Step 2. $1.0 4^{10} \approx 1.4802$ , so $\approx 7401$ .
Step 1. AP: $u_{1} = 1000$ , $d = 200$ , $u_{12} = 1000 + 11 \times 200 = 3200$ .
Step 2. $S_{12} = \frac{12}{2} (1000 + 3200) = 25200$ FRW.
Step 1. $3, 6, \dots, 99$ : 33 terms.
Step 2. $S = \frac{33}{2} (3 + 99) = 33 \times 51 = 1683$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$62$ · $- 52$ · $121$ · $2500$
1. An arithmetic sequence has $u_{1} = 5$ and $d = - 3$ . Find $u_{20}$ . →
2. Find the sum of the first 50 odd numbers. →
3. Find the sum of the first 5 terms of $81, 27, 9, \dots$ →
Show answer
1: $- 52$ · 2: $2500$ · 3: $121$
- Step 1. $u_{n} = u_{1} + (n - 1) d$ .
  Step 2. $5 + 19 \times (- 3) = 5 - 57 = - 52$ .
- Step 1. $1, 3, 5, \dots$ : AP with $u_{1} = 1$ , $d = 2$ ; $u_{50} = 99$ .
  Step 2. $S_{50} = \frac{50}{2} (1 + 99) = 2500$ .
- Step 1. $r = \frac{1}{3}$ : $S_{5} = 81 \times \frac{1 - ( 1/3 ) ^{5}}{1 - 1/3}$ .
  Step 2. $= 81 \times \frac{242}{243} \times \frac{3}{2} = 121$ (check: $81 + 27 + 9 + 3 + 1$ ).

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