Theorem

An evenly spaced run averages to its middle term

Algebra • Sequence and Series • Arithmetic Progression

Formula

a_{1} + a_{2} + \dots + a_{n} = \frac{n ( a _{1} + a _{n} )}{2}

For

n

terms in arithmetic progression, the mean is

\frac{a _{1} + a _{n}}{2}

and the sum is

\frac{n ( a _{1} + a _{n} )}{2}

; when

n

is odd the mean is exactly the middle term

a_{(n + 1) /2}

, so the sum is

n

times that middle term.

How to use it

1With an odd count, read the middle term straight off and multiply by the count - no formula needed.
2It applies to any constant step, so consecutive integers, consecutive evens and consecutive odds are all covered.
3Run it backwards: knowing the mean of an odd-length run tells you its middle term.

Worked example

Find the sum of the five consecutive odd numbers starting at

7

Step-by-step

1
$7, 9, 11, 13, 15$ with middle term $11$
Why: Five terms in arithmetic progression of step $2$ , so the third is the middle one.
2
$sum = 5 \times 11 = 55$

Answer

55

Watch out

With an even count there is no middle TERM - the mean falls between the two central terms and

\frac{a _{1} + a _{n}}{2}

must be used instead. Two separate runs cannot be merged this way: their combined mean is the weighted mean of the two sums over the TOTAL count, not the average of the two middle terms.

Also known as

sum of an AP, arithmetic progression sum, average of consecutive numbers, middle term, sum of consecutive integers

Appears in past papers

Formula

Worked example

Appears in past papers

2020 (II)

2020 (I)

2019 (I)

2018 (I)

2017 (I)