SPI Mathematics · Sequence and Series

Sum of n Terms of an AP

The sum of n terms of an AP is n/2 times the first term plus the last term, or n/2 times (2a + (n − 1)d) when the last term is not known.

All Sequence and Series notes

Why this matters

Fourteen questions in the bank, 8 EASY, 4 MODERATE and 2 HARD, and one is from the 2022 paper. Two formulas do all the work, plus three habits: knowing the standard sums, solving for n when the sum is given, and reading terms off a sum written in n. The two HARD ones are the same formulas with letters in place of numbers.

Concept 1 of 4: The two sum formulas

Write the sum forwards, then backwards under it, and add the two rows. Every pair adds up to the first term plus the last, and there are n pairs. That is twice the sum, so halve it. When the last term is not given, replace it with a + (n − 1)d.

Definition

$S_n = \dfrac{n}{2}(a + l)$ when the first term $a$ and the last term $l$ are known.
$S_n = \dfrac{n}{2}[2a + (n - 1)d]$ when $a$ and $d$ are known.
If two terms are given instead, find $a$ and $d$ first (two equations), then add.
With letters in place of numbers the formulas are the same. Check a letter answer by putting in small numbers, such as $a = 1$ and $d = 1$ .

Sum of n terms

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

athe first term
lthe last term
dthe common difference

Worked example

Find the sum of the first 20 terms of the AP 3, 7, 11, ...

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Sequence and Series · Sum of n Terms of an AP

The sum of first 51 terms of the A.P. whose 2nd term is 2 and 4th term is 8 is

Giving the last term when the sum is asked

For 1, 4, 7, ... the 10th term is 28 but the sum of 10 terms is 145. An option that matches the nth term formula is a common wrong answer.

Halving twice

The

\frac{n}{2}

already does the halving. Do not also divide

(a + l)

by 2, or the answer comes out half the right size.

Drill 3 more on the two sum formulas

Concept 2 of 4: Standard sums of natural, odd and even numbers

A few sums come up so often that it pays to know their answers by heart. Each one is just the sum formula worked out once.

Definition

First $n$ natural numbers: $1 + 2 + \cdots + n = \dfrac{n(n + 1)}{2}$ .
First $n$ odd numbers: $1 + 3 + 5 + \cdots + (2n - 1) = n^2$ .
First $n$ even numbers: $2 + 4 + 6 + \cdots + 2n = n(n + 1)$ .
First $n$ multiples of $k$ : $k + 2k + \cdots + nk = k \times \dfrac{n(n + 1)}{2}$ .

Standard sums

1 + 2 + \cdots + n = \frac{n(n + 1)}{2} \qquad 1 + 3 + \cdots + (2n - 1) = n^2 \qquad 2 + 4 + \cdots + 2n = n(n + 1)

Worked example

Find the sum of the first 12 multiples of 5.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q74Easy

Example 2 · Sequence and Series · Sum of n Terms of an AP

The sum of first fifteen multiples of 8 is ____.

Mixing up the odd and the even sums

The first 10 odd numbers add up to 100 and the first 10 even numbers to 110. Each even number is one more than the matching odd number, so the even sum is larger by

n

Using the last number as n

1 + 3 + \cdots + 41

has 21 terms, because

2n - 1 = 41

gives

n = 21

. Its sum is

21^2 = 441

, not

41^2

Drill 3 more on standard sums of natural, odd and even numbers

Concept 3 of 4: Finding the number of terms from the sum

When the sum is given and the number of terms is not, put the sum into a formula and solve for n. With the first and last terms known it is a simple equation; with only a and d known it is a quadratic.

Definition

First and last terms known: $\dfrac{n}{2}(a + l) = S$ gives $n = \dfrac{2S}{a + l}$ .
Only $a$ and $d$ known: $\dfrac{n}{2}[2a + (n - 1)d] = S$ is a quadratic in $n$ .
Keep only the positive whole-number root, because $n$ counts terms.

Number of terms from the sum

n = \frac{2S}{a + l}

Worked example

How many terms of the AP 5, 9, 13, ... add up to 230?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Sequence and Series · Sum of n Terms of an AP

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

Keeping the negative root

The quadratic for

n

usually has a negative or fractional second root. A number of terms is a positive whole number, so that root is thrown away.

Forgetting the half

Writing

n(a + l) = S

without the

\frac{1}{2}

gives half the right number of terms. The formula is

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

Drill 1 more on finding the number of terms from the sum

Concept 4 of 4: Terms from a sum written in n

If you know the sum for every number of terms, the nth term is what the nth term adds: the sum of n terms minus the sum of n minus 1 terms. The first term is simply the sum of one term.

Definition

$a_1 = S_1$ .
$a_n = S_n - S_{n-1}$ .
$d = a_2 - a_1$ , where $a_2 = S_2 - S_1$ .
Shortcut: if $S_n = pn^2 + qn$ , then $d = 2p$ and $a_1 = p + q$ .

Terms from the sum

a_1 = S_1 \qquad a_n = S_n - S_{n-1} \qquad S_n = pn^2 + qn \Rightarrow d = 2p

Worked example

The sum of the first

n

terms of an AP is

2n^2 + 3n

. Find its first term, its common difference and its 5th term.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Sequence and Series · Sum of n Terms of an AP

The sum of first n terms of an AP is

(5n - n^2)

. The nth term of the AP is

Reading the sum as a term

S_n = n^2 + 5n

, the 3rd term is not

S_3 = 24

. It is

S_3 - S_2 = 24 - 14 = 10

Expanding the bracket carelessly

S_{n-1}

means

n - 1

in place of every

n

, so

(n - 1)^2 = n^2 - 2n + 1

. The minus sign in front of

S_{n-1}

then changes every sign inside it.

Drill 3 more on terms from a sum written in n

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

The two sum formulas
Sum of n terms
$S_n = \frac{n}{2}(a + l) = \frac{n}{2}\left[2a + (n - 1)d\right]$
Standard sums of natural, odd and even numbers
Standard sums
$1 + 2 + \cdots + n = \frac{n(n + 1)}{2} \qquad 1 + 3 + \cdots + (2n - 1) = n^2 \qquad 2 + 4 + \cdots + 2n = n(n + 1)$
Finding the number of terms from the sum
Number of terms from the sum
$n = \frac{2S}{a + l}$
Terms from a sum written in n
Terms from the sum
$a_1 = S_1 \qquad a_n = S_n - S_{n-1} \qquad S_n = pn^2 + qn \Rightarrow d = 2p$

Watch out for (8)

Giving the last term when the sum is asked
→ The two sum formulas
Halving twice
→ The two sum formulas
Mixing up the odd and the even sums
→ Standard sums of natural, odd and even numbers
Using the last number as n
→ Standard sums of natural, odd and even numbers
Keeping the negative root
→ Finding the number of terms from the sum
Forgetting the half
→ Finding the number of terms from the sum
Reading the sum as a term
→ Terms from a sum written in n
Expanding the bracket carelessly
→ Terms from a sum written in n

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