Toolkit 25

Arithmetic Sequence and Series

a_n=a_1+(n-1)d

S_n=a_1+a_2+\cdots+a_n=\frac{n}{2}(a_1+a_n)

Proof

Since consecutive terms differ by $d$ ,

a_2=a_1+d,\qquad a_3=a_1+2d,

and, in general,

a_n=a_1+(n-1)d.

Now let

S_n=a_1+a_2+\cdots+a_{n-1}+a_n.

Writing the same sum in reverse order gives

S_n=a_n+a_{n-1}+\cdots+a_2+a_1.

Adding the two equations term by term,

2S_n = (a_1+a_n)+(a_2+a_{n-1})+\cdots+(a_n+a_1).

Every pair is equal to $a_1+a_n$ , and there are $n$ pairs. Hence,

2S_n=n(a_1+a_n).

Therefore,

S_n=\frac{n}{2}(a_1+a_n). \quad\square