Question

If $S_1$ be the sum of (2n + 1) terms of an A.P. and $S_2$ be the sum of its odd terms, then prove that $S_1:S_2=(2n+1):(n+1)$.

Answer 1

In A.P. ‘a’ be the first term and d be the common difference then
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
Given that $S_1$ be the sum of (2n + 1) terms then
$S_1=\frac{(2n-1)}{2}[2a+(2n+1-1\left)d\right]$
$S_1=\frac{(2n-1)}{2}[2a+2(a+nd\left)\right]$
$S_1=(2n+1)(a+nd)$
$S_2$ be the sum of its odd terms i.e.
$S_2=a_1+a_3+a_5+...+a_{(}2n-1)$
$S_2=a+(a+2d)+...+(a+2nd)$
$S_2=(n+1)a+2\frac{n(n+1)}{2}d$
$\therefore S_2=(n+1)a+n(n+1)d$
$S_2=(n+1)[a+nd]$
$S_1:S_2=(2n+1)(a+nd):(n+1)(a+nd)$
$S_1:S_2=(2n+1):(n+1)$