Sums of Arithmetic Progressions

Pure Mathematics 1

Mathematics9709

Pair the first term with the last

In an AP, the first and last terms have the same total as the second and second-last terms. Write the whole sum forwards and backwards:

Forwards	Backwards	Pair sum
2	14	16
5	11	16
8	8	16
11	5	16
14	2	16

2S₅ = 5 × 16 = 80

S₅ = 40

Two copies of the sum give five equal pairs.

Let $l$ be the last term. Adding the two rows gives $n$ pairs of $a+l$ . Each original term was used twice, so divide by two:

S_n=\frac n2(a+l)

This still works for an odd number of terms: the middle term pairs with itself across the two rows. Substituting $l=a+(n-1)d$ gives

S_n=\frac n2[2a+(n-1)d]

Use the information you have

Use the first form when you know both end terms; use the second when you know $a$ and $d$ . In either case, $n$ counts terms, not changes.

Worked example

Find the sum of the first 20 terms of $5,11,17,\ldots$ .

Here $a=5$ , $d=6$ and $n=20$ .

S_{20}=\frac{20}{2}[10+19(6)]=1240

Check: the last term is 119, so the average term is $(5+119)/2=62$ . Twenty terms give $20(62)=1240$ .

When a term and a sum are given

Worked example

An AP has $u_6=17$ and $S_4=26$ . Find its first term and difference.

A term gives one equation. A sum gives another:

a+5d=17

\begin{aligned}2(2a+3d)&=26\\2a+3d&=13\end{aligned}

Double the first equation, then subtract the second. This gives $7d=21$ , so $d=3$ and $a=2$ . The first four terms are 2, 5, 8, 11; their sum is 26.

Common mistake

S_4=26

does not mean the fourth term is 26. Write the sum equation before solving.

Pair the first term with the last

In an AP, the first and last terms have the same total as the second and second-last terms. Write the whole sum forwards and backwards:

Forwards	Backwards	Pair sum
2	14	16
5	11	16
8	8	16
11	5	16
14	2	16

2S₅ = 5 × 16 = 80

S₅ = 40

Two copies of the sum give five equal pairs.

Let $l$ be the last term. Adding the two rows gives $n$ pairs of $a+l$ . Each original term was used twice, so divide by two:

S_n=\frac n2(a+l)

This still works for an odd number of terms: the middle term pairs with itself across the two rows. Substituting $l=a+(n-1)d$ gives

S_n=\frac n2[2a+(n-1)d]

Use the information you have

Use the first form when you know both end terms; use the second when you know $a$ and $d$ . In either case, $n$ counts terms, not changes.

Worked example

Find the sum of the first 20 terms of $5,11,17,\ldots$ .

Here $a=5$ , $d=6$ and $n=20$ .

S_{20}=\frac{20}{2}[10+19(6)]=1240

Check: the last term is 119, so the average term is $(5+119)/2=62$ . Twenty terms give $20(62)=1240$ .

When a term and a sum are given

Worked example

An AP has $u_6=17$ and $S_4=26$ . Find its first term and difference.

A term gives one equation. A sum gives another:

a+5d=17

\begin{aligned}2(2a+3d)&=26\\2a+3d&=13\end{aligned}

Double the first equation, then subtract the second. This gives $7d=21$ , so $d=3$ and $a=2$ . The first four terms are 2, 5, 8, 11; their sum is 26.

Common mistake

S_4=26

does not mean the fourth term is 26. Write the sum equation before solving.

Pair the first term with the last

Use the information you have

When a term and a sum are given

Now you try

Sums of Arithmetic Progressions

Pair the first term with the last

Use the information you have

When a term and a sum are given

Now you try