Question

If there are $(2n+1)$ terms in an AP, then show that : $\frac{Sumofoddterms}{Sumofeventerms}=\frac{n+1}{n}$.

Answer 1

Let $a$ be the first term and $d$ be the common difference of the given AP.

Now, sum of odd terms = $S_1=a_1+a_3+a_5+.....+a_{2n+1}$

$S_1=\frac{n+1}{2}(a_1+a_{2n+1})$

$S_1=\frac{n+1}{2}[a_1+a_1+(2n+1-1\left)d\right]$

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

Also, sum of even terms = $S_2=a_2+a_4+a_6+....+a_{2n}$

$S_2=\frac{n}{2}(a_2+a_{2n})$

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

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

Therefore,

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