Sequences and arithmetic progressions · Sequences & series (AP, GP)

A sequence is a list of numbers in order, such as $2, 5, 10, 17, \dots$ Each number is a term. The first term is $T_1$ , the second $T_2$ , and so on. The $n$ th term (or general term) $T_n$ is a formula in $n$ that gives any term: put in $n = 1, 2, 3, \dots$ and out come the terms in order.

T_n = n^2 + 1 \quad\text{gives}\quad 2,\ 5,\ 10,\ 17,\ \dots

The nth term is a rulePut in n, get Tₙ: here Tₙ = n² + 1

Arithmetic progressions

In an arithmetic progression (A.P., or linear sequence) you add the same number every time. That number is the common difference $d$ , found by taking any term from the next one: $d = T_2 - T_1$ . With first term $a$ :

a,\quad a + d,\quad a + 2d,\quad a + 3d,\ \dots

The first term has no $d$ , the second has one, the third has two. So the $n$ th term has $n - 1$ of them:

T_n = a + (n - 1)d

nth term of an A.P.Tₙ = a + (n − 1)d: T₅ has 4 blocks of d

Try it

The terms of an A.P.Set a and d, then pick the terms

13T63 + 5 × 2a + (n − 1)d5steps of d from T₁

a = 3d = 2n = 6

The sequence rises by the same amount each time: d = 2. To reach T6 you start at a = 3 and take 5 steps of d, one fewer than n, so T6 = 3 + 5 × 2 = 13.

In “nth term” mode, the gold term is always $n - 1$ steps of $d$ from the first, never $n$ . A negative $d$ makes the terms fall. Then switch to “From two terms”: the gap between two terms is a whole number of steps of $d$ , whatever the first term is.

Finding a and d from two terms

Most exam questions give two terms and ask for $a$ and $d$ . Write each term as $a + (n - 1)d$ , then subtract the equations: $a$ cancels and leaves $d$ . The gap between $T_p$ and $T_q$ is $q - p$ steps of $d$ .

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q2

The second, fourth and sixth terms of an Arithmetic Progression (A.P.) are $x - 1$ , $x + 1$ and $7$ respectively. Find the:

common difference;

first term;

value of $x$ .

Write the terms
$T_2 = a + d = x - 1$ , $\quad T_4 = a + 3d = x + 1$ , $\quad T_6 = a + 5d = 7$ .
Think first. Write the 2nd, 4th and 6th terms in terms of a and d.
(a) Subtract two of them
Take $T_2$ from $T_4$ : $(a + 3d) - (a + d) = (x + 1) - (x - 1)$ , so $2d = 2$ and the common difference is $d = 1$ .
Think first. Which two equations make both a and x cancel?
(b) The first term
$a + 5d = 7$ , so $a + 5 = 7$ and the first term is $a = 2$ .
Think first. Which equation has no x in it?
(c) The value of x
$a + d = x - 1$ : $3 = x - 1$ , so $x = 4$ . Check with $T_4$ : $a + 3d = 5$ and $x + 1 = 5$ . ✓
Think first. Now use T₂.

How many terms?

When you know the last term $l$ , put $T_n = l$ and solve for $n$ .

Three consecutive terms

If $x$ , $y$ and $z$ are consecutive terms of an A.P., the gaps are equal: $y - x = z - y$ . So

2y = x + z

The middle term is the average (the arithmetic mean) of its neighbours.

Consecutive terms of an A.P.y − x = z − y, so 2y = x + z

Worked example · WAEC 2020

WAEC 2020 · Paper 2 · Q6 (a)

Given that $(y + 2)$ , $(y + 3)$ and $(2y^2 + 1)$ are consecutive terms of an Arithmetic Progression (A.P.), find the possible values of $y$ .

Set the gaps equal
$(y + 3) - (y + 2) = (2y^2 + 1) - (y + 3)$
Think first. What is true about the two gaps between consecutive terms?
Simplify each side
$1 = 2y^2 - y - 2$
Think first. The left side is a number. What is it?
Solve the quadratic
$2y^2 - y - 3 = 0$ , which factorises as $(2y - 3)(y + 1) = 0$ . So $y = \frac32$ or $y = -1$ .
Think first. Make one side 0, then factorise.
Check both
$y = -1$ gives $1, 2, 3$ and $y = \frac32$ gives $3\frac12, 4\frac12, 5\frac12$ . Both have equal gaps, so both values stand.
Think first. Do both values give an A.P.?

Your turn

WAEC 2021 · Paper 2 · Q7 (b)

(b)
The 7th and 18th terms of an Arithmetic Progression (A.P.) are $15\frac14$ and $42\frac34$ respectively. Find the: (i) $n$ th term; (ii) sixth term, of the progression.

Worked solution (try it first)

(b)(i)

The $n$ th term is $U_n = a + (n - 1)d$ .
So $a + 6d = 15\frac14 = \frac{61}{4}$ and $a + 17d = 42\frac34 = \frac{171}{4}$ .
Take the first from the second: $11d = \frac{110}{4}$ , so $d = \frac{10}{4} = \frac52$ .
Then $a = \frac{61}{4} - 6 \times \frac52$
$= \frac{61}{4} - 15$
$= \frac14$ .
So $U_n = \frac14 + \frac52(n - 1)$
$= \frac{1 + 10n - 10}{4}$
$= \frac{10n - 9}{4}$ .

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