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Mathematics and Statistics · Ch 4 — Sequences and Series

Sum of n Terms of an AP

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Sum of n Terms of an AP

The sum of the first nn terms of an AP, SnS_n, has a direct closed-form formula — useful whenever a running total is needed (total savings over several months, total output over several years).

Note

Sum-to-n-Terms of an AP

Sn=n2[ 2a+(n−1)d ]=n2(a+l)S_n = \frac{n}{2}\big[\,2a + (n-1)d\,\big] = \frac{n}{2}(a + l)

where l=tnl = t_n is the last (i.e. nn-th) term. The second form is convenient whenever the last term is already known.

Where it comes from (the pairing idea): write the sum forwards and backwards and add — the first and last term give a+la + l, the second and second-last also give a+la + l, and so on. There are nn such equal pairs across the two copies, so 2Sn=n(a+l)2S_n = n(a+l), giving Sn=n2(a+l)S_n = \tfrac{n}{2}(a+l). (This is the reasoning famously attributed to the young Gauss.)

Example: the sum of the first 2020 terms of 4,7,10,…4, 7, 10, \ldots (a=4, d=3a = 4,\ d = 3) is

S20=202[2(4)+19(3)]=10(8+57)=10(65)=650.S_{20} = \frac{20}{2}\big[2(4) + 19(3)\big] = 10(8 + 57) = 10(65) = 650.

Checked by the other form: the 2020th term is l=4+19(3)=61l = 4 + 19(3) = 61, so S20=202(4+61)=10(65)=650S_{20} = \tfrac{20}{2}(4 + 61) = 10(65) = 650 — matches.

Note

A Useful Link …

Definition 5Sum to n terms (Sn) of an AP

Sn=n2[2a+(n−1)d]=n2(a+l)S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l), the total of the first nn terms of an AP, where $l = t_ …