The sum of an arithmetic progression · Sequences & series (AP, GP)

When you add the terms of a sequence you get a series. $S_n$ stands for the sum of the first $n$ terms. For an A.P. there’s a shortcut. Write the sum forwards, then write it again backwards underneath. Each pair, one above the other, adds to the first term plus the last term, $a + l$ . There are $n$ pairs, so two copies of the sum make $n(a + l)$ , and

S_n = \frac{n}{2}(a + l)

The last term is $l = a + (n - 1)d$ . Put that in when you don’t know $l$ :

S_n = \frac{n}{2}\big[2a + (n - 1)d\big]

Sum of an A.P.Two copies make n columns of a + l, so Sₙ = n/2 (a + l)

Try it

Adding up an A.P.Set a, d and n, then turn the copy round

2 + 17 = 19a + l, each column6 × 19 = 114n(a + l), both copies57S6, half of that

a = 2d = 3n = 6

Green is 2 + 5 + 8 + 11 + 14 + 17. Gold is the same terms the other way round. Each column is a first-and-last pair adding to a + l = 19, and there are 6 columns, so two copies of the sum make 6 × 19 = 114. One copy is half: S6 = 6/2 × 19 = 57.

Whatever $a$ , $d$ and $n$ you choose, the columns all come out the same height, $a + l$ . That’s why the sum is $n$ columns of $a + l$ , halved.

Using the formula

Use $\frac{n}{2}(a + l)$ when you know the last term, and $\frac{n}{2}[2a + (n - 1)d]$ when you know $d$ . Either way you need $n$ , the number of terms.

A term and a sum given

When a question gives one term and one sum, write two equations: one from $T_n = a + (n - 1)d$ and one from $S_n = \frac{n}{2}[2a + (n - 1)d]$ . Then solve them together, as in simultaneous equations.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q1

The third term of an Arithmetic Progression (A.P.) is 23 and the sum of the first seven terms is 210. Find the:

common difference;

first term;

sum of the first 20 terms.

Equation from the term
$T_3 = a + 2d = 23$
Think first. Write the third term in terms of a and d.
Equation from the sum
$S_7 = \frac72(2a + 6d) = 7(a + 3d) = 210$ , so $a + 3d = 30$ .
Think first. Put n = 7 into the sum formula and simplify.
(a) Subtract
$(a + 3d) - (a + 2d) = 30 - 23$ , so the common difference is $d = 7$ .
Think first. Which letter cancels?
(b) The first term
$a = 23 - 14 = 9$ .
Think first. Put d = 7 into a + 2d = 23.
(c) The sum of 20 terms
$S_{20} = \frac{20}{2}[2 \times 9 + 19 \times 7] = 10 \times (18 + 133) = 10 \times 151 = 1510$ .
Think first. Now you know a and d. Which formula?

How many terms make a sum?

When the sum is given and $n$ is unknown, the sum formula becomes a quadratic equation in $n$ . $n$ counts terms, so keep only a positive whole number.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q6

The heights of students in a class form an Arithmetic Progression (A.P.) such that the tallest student is $1.90\text{ m}$ . The common difference in their heights is $0.05\text{ m}$ and the sum of their heights is $23.25\text{ m}$ .

How many students are in the class?

Find the height of the shortest student.

Name what you know
Take the heights from shortest to tallest: first term $a$ (unknown), $d = 0.05$ , last term $l = 1.90$ , and $n$ students (unknown).
Think first. The heights go up by 0.05 m. Is the tallest student the first term or the last?
Equation from the last term
$1.90 = a + 0.05(n - 1)$ , so $a = 1.95 - 0.05n$ .
Think first. Write l = a + (n − 1)d and make a the subject.
Equation from the sum
$\frac{n}{2}(a + 1.90) = 23.25$ , so $n(1.95 - 0.05n + 1.90) = 46.5$ , which is $n(3.85 - 0.05n) = 46.5$ .
Think first. Use Sₙ = n/2 (a + l) and put in a from the last step.
(a) Solve the quadratic
$3.85n - 0.05n^2 = 46.5$ . Times 20: $77n - n^2 = 930$ , so $n^2 - 77n + 930 = 0$ and $(n - 15)(n - 62) = 0$ .
Think first. Multiply by 20 to clear the decimals.
Choose the right n
$n = 62$ gives $a = 1.95 - 3.10 = -1.15$ m, which is impossible. So there are 15 students.
Think first. What shortest height does each value give?
(b) The shortest student
$a = 1.95 - 0.05 \times 15 = 1.95 - 0.75 = 1.20$ m. Check: $\frac{15}{2}(1.20 + 1.90) = 23.25$ ✓.
Think first. Put n = 15 into a = 1.95 − 0.05n.

From the sum back to a term

If you’re given a formula for $S_n$ , any term is the difference of two sums: $T_n = S_n - S_{n - 1}$ . The first $n$ terms minus the first $n - 1$ terms leaves the $n$ th.

Your turn

WAEC 2018 · Paper 2 · Q1

Sequences & series (AP, GP)

The sum of the first ten terms of an Arithmetic Progression (A.P.) is 130. If the fifth term is 3 times the first term, find the:

(a)
common difference;
(b)
first term;
(c)
number of terms of the A.P. if the last term is 28.

Worked solution (try it first)

Write each fact as an equation.
The sum of the first ten terms: $S_{10} = \frac{10}{2}[2a + 9d]$
$= 5(2a + 9d)$
$= 130$ , so $2a + 9d = 26$ .
The fifth term is 3 times the first: $a + 4d = 3a$ , so $4d = 2a$ and $a = 2d$ .

(a)

Put $a = 2d$ into $2a + 9d = 26$ : $4d + 9d = 26$ , so $13d = 26$ and the common difference is $d = 2$ .

(b)

The first term is $a = 2d = 4$ .
(Check: $5(8 + 18) = 130$ ✓ and $T_5 = 4 + 8 = 12 = 3 \times 4$ ✓.)

(c)

The last term is 28: $a + (n - 1)d = 28$ , so $4 + 2(n - 1) = 28$ , $2(n - 1) = 24$ , $n - 1 = 12$ and there are $n = 13$ terms.

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