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

Let the A.P. be a, a + d, a + 2d
$S_{(2n+1)}=S_1=\frac{(2n+1)}{2}[2a+(2n+1-1\left)d\right]S_1=\frac{(2n+1)}{2}[2a+2nd]=(2n+1)(a+nd)\dots \dots \left(i\right)$
Sum of odd terms $=S_2$
$S_2=\frac{(n+1)}{2}[2a+(n+1-1\left)\right(2d\left)\right][2a+(2n+1-1\left)d\right]S_1=\frac{(2n+1)}{2}[2n+2nd]=(2n+1)(a+nd)\dots \dots \left(i\right)$
Sum of odd terms $=S_2$
$S_2=\frac{(n+1)}{2}[2a+(n+1-1\left)\right(2d\left)\right]=\frac{(n+1)}{2}[2a+2nd]S_2=(n+1)(a+nd)\dots \dots \left(ii\right)$
From equation (i) and (ii),
$S_1:S_2=(2n+1)(a+nd):(n+1)(a+nd)S_1:S_2=(2n+1);(n+1)$