Arithmetic Progression: nth Term and Sum | NIMCET

Quick answer: An arithmetic progression (AP) has a constant difference d between terms. With first term a, the nth term is $a_n = a + (n - 1)d$ , and the sum of n terms is $S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}(a + l)$ , where l is the last term. The arithmetic mean of a and b is $\frac{a + b}{2}$ , and $a_n = S_n - S_{n - 1}$ .

Key formulas

Quantity	Formula
Common difference	$d = a_2 - a_1$
nth term	$a_n = a + (n - 1)d$
nth term from the end	$l - (n - 1)d$
Sum of n terms	$S_n = \frac{n}{2}[2a + (n - 1)d]$
Sum using last term	$S_n = \frac{n}{2}(a + l)$
nth term from sums	$a_n = S_n - S_{n - 1}$

Test for an AP: a sequence is an AP if and only if $a_n$ is linear in n, or equivalently $S_n$ is a quadratic in n with no constant term.

Arithmetic mean

The AM of a and b is $A = \frac{a + b}{2}$ .
Three numbers a, b, c are in AP if and only if $2b = a + c$ .

Inserting n arithmetic means between a and b: the common difference is

$d = \frac{b - a}{n + 1}$

and the sum of the n means is $n \cdot \frac{a + b}{2}$ .

Properties of an AP

Property	Statement
Adding a constant	Still an AP, same d
Multiplying by k	Still an AP, difference kd
Equidistant terms	$a_1 + a_n = a_2 + a_{n - 1} = \ldots$
Choosing 3 terms	$a - d$ , $a$ , $a + d$
Choosing 4 terms	$a - 3d$ , $a - d$ , $a + d$ , $a + 3d$

Sums of natural numbers

Sum	Formula
$\sum n$	$\frac{n(n + 1)}{2}$
$\sum n^2$	$\frac{n(n + 1)(2n + 1)}{6}$
$\sum n^3$	$\left[\frac{n(n + 1)}{2}\right]^2$
Sum of first n odd numbers	$n^2$

Worked examples

Example 1: nth term

Find the 20th term of 7, 11, 15, ….

$a = 7$ , $d = 4$ : $a_{20} = 7 + 19 \times 4 = 83$ .

Example 2: sum of terms

Find the sum of all two-digit multiples of 3.

Step 1: The terms are 12, 15, …, 99, with $d = 3$ .

Step 2: $99 = 12 + (n - 1)3$ , so $n = 30$ .

Step 3: $S = \frac{30}{2}(12 + 99) = 1665$ .

Example 3: term from the sum

If $S_n = 3n^2 + 2n$ , find the nth term.

$a_n = S_n - S_{n - 1} = 3n^2 + 2n - [3(n - 1)^2 + 2(n - 1)] = 6n - 1$ .

Check: $a_1 = 5 = S_1$ ✓.

Example 4: three terms in AP

Three numbers in AP have sum 15 and product 80. Find them.

Step 1: Take $a - d$ , $a$ , $a + d$ . Then $3a = 15$ , so $a = 5$ .

Step 2: $5(25 - d^2) = 80$ , so $d^2 = 9$ and $d = \pm 3$ .

Answer: 2, 5, 8.

Example 5: ratio of sums

The sums of n terms of two APs are in the ratio $(3n + 1) : (2n + 5)$ . Find the ratio of their 5th terms.

Step 1: The ratio of the kth terms equals the ratio of sums with $n = 2k - 1$ .

Step 2: For $k = 5$ , $n = 9$ : $\frac{28}{23}$ .

Summary table

Idea	Key result
nth term	$a + (n - 1)d$
Sum	$\frac{n}{2}[2a + (n - 1)d]$
a, b, c in AP	$2b = a + c$
Ratio of kth terms from sums	Put $n = 2k - 1$
$\sum n^2$	$\frac{n(n + 1)(2n + 1)}{6}$

Exam tips

Choose symmetric terms ( $a - d$ , $a$ , $a + d$ ) whenever the sum is given.
If $S_n$ is given as a formula, the common difference is twice the coefficient of $n^2$ .
For the ratio of kth terms of two APs, substitute $n = 2k - 1$ in the ratio of sums.

Next, read about geometric progressions and the infinite GP sum, or see the NIMCET Algebra chapter.

Common mistakes

Using $a + nd$ instead of $a + (n - 1)d$ for the nth term.
Miscounting the number of terms in a range.
Using $S_n - S_{n - 1}$ for $n = 1$ ; there $a_1 = S_1$ .
Choosing $a, a + d, a + 2d$ when symmetric terms would simplify the algebra.

Frequently Asked Questions

What is the nth term of an AP?

aₙ = a + (n − 1)d, where a is the first term and d the common difference.

What is the sum of n terms of an AP?

Sₙ = (n/2)[2a + (n − 1)d], or (n/2)(first term + last term).

What is the arithmetic mean of two numbers?

The AM of a and b is (a + b)/2.

How do you find the nth term from Sₙ?

aₙ = Sₙ − Sₙ₋₁ for n ≥ 2, and a₁ = S₁.

What is the sum of the squares of the first n natural numbers?

n(n + 1)(2n + 1)/6.