Sum of an Arithmetic Series (Edexcel IGCSE Maths A (Modular))

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Written by: Mark Curtis

Reviewed by: Dan Finlay

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Sum of an Arithmetic Series

What is the sum of an arithmetic sequence?

The sum of an arithmetic sequence (a series) means the terms in an arithmetic sequence are added together
- For example, the sum of the first 5 terms in the sequence 2, 4, 6, 8, 10, 12, ... is
  - 2 + 4 + 6 + 8 + 10 = 30

What is the formula for the sum of an arithmetic sequence?

The formula for the sum of the first $n$ terms in an arithmetic sequence is
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
- $a$ is the first term
- $d$ is the common difference
- $n$ is the number of terms being added
- $S_{n}$ is the sum of the first $n$ terms

How do I use the formula for the sum of an arithmetic sequence?

You may have to substitute values into the formula to find $S_{n}$
- For example, if $a = 2$ and $d = 5$ then the sum of the first ten terms is $S_{10}$
  - Substitute $a = 2$ , $d = 5$ and $n = 10$ into the formula
You may have to form equations in terms of $a$ and $d$ when given information about the sum of the first $n$ terms
- This may lead to simultaneous equations
  - The second equation could come from the formula for the $n$ ^th term, $u_{n} = a + (n - 1) d$
You may have to form a quadratic equation or quadratic inequality in $n$
- When solving, remember that $n$ must be a positive integer

Examiner Tips and Tricks

The formula $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ is given in the formula booklet.

Worked Example

An arithmetic sequence is given by 6, 13, 20, 27, ...

(a) Find the sum of the first twenty terms.

The formula for the sum of the first $n$ terms in an arithmetic sequence is given by $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Here, $a = 6$ , $d = 7$ and $n = 20$ , so substitute these in

$\begin{array}{rcl} S_{20} & = & \frac{20}{2} [2 \times 6 + (20 - 1) \times 7] \\ = & 10 (12 + 19 \times 7) \\ = & 1450 \end{array}$

The sum of the first twenty terms is 1450

(b) What is the minimum number of terms required for the sum of the first $n$ terms to exceed 4000?

This question does not tell you the value of $n$
You know $a = 6$ and $d = 7$ from before
Use the formula for the sum of the first $n$ terms to write $S_{n} > 4000$

$\frac{n}{2} [2 \times 6 + (n - 1) \times 7] > 4000$

Multiply both sides by 2 and simplify

$\begin{array}{rcl} n [12 + 7 (n - 1)] & > & 8000 \\ n [12 + 7 n - 7] & > & 8000 \\ n (5 + 7 n) & > & 8000 \\ 5 n + 7 n^{2} & > & 8000 \end{array}$

This is a quadratic inequality
Bring all the terms to the left-hand side

$7 n^{2} + 5 n - 8000 > 0$

If you know how to solve a quadratic inequality, you can use that method
If not, make the inequality an equals sign and solve (e.g. the quadratic formula)

$n = \frac{- 5 \pm \sqrt{5^{2} - 4 \times 7 \times (- 8000)}}{14}$

This gives

$n = 33.450913 . . .$ or $n = - 34.165199 . . .$

$n$ is the number of terms so cannot be negative; it must be near 33
Test $n = 33$ to see if $S_{n} > 4000$ from the original question

$S_{33} = \frac{33}{2} [2 \times 6 + (33 - 1) \times 7] = 3894 < 4000$

This is not enough terms for the sum to exceed 4000
Test $n = 34$ to see if $S_{n} > 4000$

$S_{34} = \frac{34}{2} [2 \times 6 + (34 - 1) \times 7] = 4250 > 4000$

This is the first time the sum exceeds 4000

You need a minimum of 34 terms for the sum to exceed 4000

Last updated: 18 October 2024

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