Sum of an Arithmetic Series (Edexcel IGCSE Maths)

Spec point — 3.1C

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

Exam Hint 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.

Answer:

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

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

Final answer: 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?

Answer:

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

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

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

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

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

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