Arithmetic sequences and their sums – Sequences (Senior 5 Mathematics) | Ganzaa

Sequences

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Senior 5 · Mathematics

Sequences

4 lessons

Lesson 2 of 4

Arithmetic sequences and their sums

25 min

Saving a little more each week

A member of a savings group (an ikimina) puts in 2,000 Frw in the first week and decides to add 500 Frw more each week than the week before: 2,000, 2,500, 3,000, 3,500, … Frw (illustrative figures). How much will she put in during week 20, and how much will she have saved after 20 weeks? Writing all twenty numbers works, but it does not work for 200. The amounts go up by the same step each time, and that one fact gives formulae for any term and for the total.

Definition

Arithmetic sequence (arithmetic progression, AP): a sequence in which each term is the previous term plus a fixed number $d$ :

u_{n+1} = u_n + d \quad \text{for every } n.

Common difference $d$ : the fixed difference $u_{n+1} - u_n$ .

Example: $20, 17, 14, 11, \dots$ is arithmetic with $d = - 3$ ; $1, 4, 9, 16, \dots$ is not, since its differences $3, 5, 7$ change.

The general term

From $u_{1}$ to $u_{n}$ there are $n - 1$ steps, each adding $d$ :

u_n = u_1 + (n - 1)d, \qquad \text{and more generally} \qquad u_n = u_p + (n - p)d.

An arithmetic sequence with $d \gt 0$ is strictly increasing, with $d \lt 0$ strictly decreasing. Its dot graph lies on a straight line of gradient $d$ , which is why the S3 linear functions and arithmetic sequences behave alike.

Three terms in a row. If $a, b, c$ are consecutive terms, then $b - a = c - b$ , so

b = \frac{a + c}{2}:

the middle term is the arithmetic mean of its neighbours.

Definition

Arithmetic mean of $a$ and $c$ : the number $\frac{a + c}{2}$ .

Inserting arithmetic means: to insert $k$ arithmetic means between $a$ and $b$ is to find $k$ numbers that make $a, \dots, b$ an arithmetic sequence of $k + 2$ terms. Then $b = a + (k + 1) d$ , so $d = \frac{b - a}{k + 1}$ .

Example: one arithmetic mean between 6 and 20 is 13.

Worked example 1: two terms fix the sequence

The 5th term of an arithmetic sequence is 17 and the 12th term is 45. Find $u_{1}$ , $d$ and the general term.

Solution $u_{12} = u_5 + 7d$ , so $45 = 17 + 7 d$ and $d = 4$ . Then $u_{1} = u_{5} - 4 d = 17 - 16 = 1$ , and

u_n = 1 + (n - 1)\times 4 = 4n - 3.

Check: $u_{12} = 48 - 3 = 45$ . ✓

Worked example 2: arithmetic means

Insert four arithmetic means between 7 and 32.

Solution Here $k = 4$ , so there are 6 terms and $d = \frac{32 - 7}{4 + 1} = 5$ . The means are $12, 17, 22, 27$ , giving $7, 12, 17, 22, 27, 32$ .

The sum of the first $n$ terms

Write the sum forwards and backwards and add the two lines term by term:

\begin{aligned} S_n &= u_1 + (u_1 + d) + \cdots + (u_n - d) + u_n \\ S_n &= u_n + (u_n - d) + \cdots + (u_1 + d) + u_1 \\ 2S_n &= (u_1 + u_n) + (u_1 + u_n) + \cdots + (u_1 + u_n) = n(u_1 + u_n). \end{aligned}

Every column adds to the same total $u_{1} + u_{n}$ , and there are $n$ columns. Figure 1 shows the same argument as a picture.

Figure 1. Two copies of an arithmetic sum, one upside down, fill a rectangle of width

n

and height

u_{1} + u_{n}

. So

2 S_{n} = n (u_{1} + u_{n})

Important

Sum of an arithmetic sequence

S_n = u_1 + u_2 + \cdots + u_n = \frac{n}{2}\left(u_1 + u_n\right) = \frac{n}{2}\left[2u_1 + (n - 1)d\right].

Use the first form when you know the last term, the second when you know $d$ . In words: the number of terms times the average of the first and last.

Worked example 3: the savings group

For the savings in the opening ( $u_{1} = 2000$ Frw, $d = 500$ Frw), find the amount saved in week 20 and the total after 20 weeks.

Solution $u_{20} = 2000 + 19 \times 500 = 11\,500$ Frw, and

S_{20} = \frac{20}{2}\left(2000 + 11\,500\right) = 10 \times 13\,500 = 135\,000 \text{ Frw}.

Worked example 4: counting the terms first

Find the sum of all the multiples of 7 between 100 and 500.

Solution The first is $105 = 7 \times 15$ and the last is $497 = 7 \times 71$ . The number of terms is $71 - 15 + 1 = 57$ (or $\frac{497 - 105}{7} + 1 = 57$ ). So

S = \frac{57}{2}(105 + 497) = 57 \times 301 = 17\,157.

Worked example 5: how many terms?

How many terms of $3 + 7 + 11 + \cdots$ are needed for the sum to exceed 1000?

Solution $u_{1} = 3$ , $d = 4$ : $S_n = \frac{n}{2}\left[6 + 4(n - 1)\right] = n(2n + 1)$ . We need $2n^2 + n - 1000 \gt 0$ . The positive root of $2 n^{2} + n - 1000 = 0$ is $n = \frac{-1 + \sqrt{8001}}{4} \approx 22.1$ , so $n \ge 23$ . Check: $S_{22} = 22 \times 45 = 990$ and $S_{23} = 23 \times 47 = 1081$ . 23 terms are needed.

Warning

Off by one. The $n$ th term has $n - 1$ steps, not $n$ : $u_{20} = u_1 + 19d$ . And the number of terms from $a$ to $b$ in steps of $d$ is $\frac{b - a}{d} + 1$ : there are 57 multiples of 7 from 105 to 497, not 56.

Tip

Sum given, term wanted. If you know $S_{n}$ for every $n$ , then $u_n = S_n - S_{n-1}$ for $n \ge 2$ , and $u_{1} = S_{1}$ .

Khan Academy: Introduction to arithmetic sequences. Watch how the common difference is found from consecutive terms, and how it leads to the formula for the nth term.

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Practice

The last two are harder.

Find the 25th term of $8, 5, 2, \dots$ .
Which term of $4, 11, 18, \dots$ is 368?
In an arithmetic sequence $u_{3} = 10$ and $u_{8} = 35$ . Find $u_{1}$ , $d$ and $u_{n}$ .
Insert three arithmetic means between $- 2$ and 18.
Find the sum of the first 40 terms of $2 + 5 + 8 + \cdots$ .
Find the sum of all the odd numbers from 1 to 199.
$x - 1$ , $2 x + 1$ and $4 x - 1$ are consecutive terms of an arithmetic sequence. Find $x$ and the three terms.
The sum of the first $n$ terms of a sequence is $S_{n} = 3 n^{2} + 2 n$ . Show that the sequence is arithmetic, and find $u_{1}$ and $d$ .

Show answersHide answers

$d = - 3$ : $u_{25} = 8 + 24(-3) = -64$ .
$4 + 7 (n - 1) = 368$ , so $n - 1 = 52$ : the 53rd term.
$5 d = 25$ , $d = 5$ , $u_{1} = 10 - 10 = 0$ , $u_{n} = 5 n - 5$ .
$d = \frac{18 - (-2)}{4} = 5$ : $3, 8, 13$ .
$S_{40} = 20\,(4 + 39 \times 3) = 20 \times 121 = 2420$ .
100 terms: $S = \frac{100}{2}(1 + 199) = 10\,000$ .
$2 (2 x + 1) = (x - 1) + (4 x - 1)$ , so $4 x + 2 = 5 x - 2$ and $x = 4$ : the terms are $3, 9, 15$ .
$u_n = S_n - S_{n-1} = 3n^2 + 2n - 3(n - 1)^2 - 2(n - 1) = 6n - 1$ for $n \ge 2$ , and $S_{1} = 5 = 6 (1) - 1$ . So $u_{n} = 6 n - 1$ : arithmetic with $u_{1} = 5$ , $d = 6$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples (the savings figures are illustrative), the practice items and the quiz questions.
Video: Introduction to arithmetic sequences, Khan Academy, embedded from YouTube.

1Saving a little more each week

2The general term

3The sum of the first n n n terms

Saving a little more each week

The general term

The sum of the first $n$ terms