Question

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

Answer 1

Let a be the first term and d be the common difference of the given A.P

Then $a_k=a+(k-1)d$

Let $S_1$ denote the sum of the (2n+1) terms and odd terms of the given A.P, Hence

$S_1=\frac{2n+1}{2}[2a+(2n+1-1\left)d\right]$ [sum of n terms of A.P = $\frac{n}{2}(2a+(n-1\left)d\right)]$

$S_1=\frac{2n+1}{2}[2a+2nd]$

$S_1=\frac{2n+1}{2}2(a+nd)$

$S_1=(2n+1)(a+nd)$

Now $S_2=a_1+a_3+a_5+.......+a_{2n+1}$

$S_2=\frac{(n+1)}{2}(a_1+a_{2n+1})$

$S_2=\frac{(n+1)}{2}.(a+a+(2n+1-1\left)d\right)$

$S_2=\frac{(n+1)}{2}.(2a+2nd)$

$S_2=\frac{(n+1)}{2}2(a+nd)$

$S_2=(n+1)(a+nd)$

$\frac{S_1}{S_2}=\frac{(2n+1)(a+nd)}{(n+1)(a+nd)}$

$\frac{S_1}{S_2}=\frac{(2n+1)}{n+1}$

$S_1:S_2=(2n+1):(n+1)$ Hence proved!