Question

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

Answer 1

Let the A.P be $x_1,x_2,...x_n,...x_{2n+1}$

and let the first term $x_1=a$ and the common difference be $d$

We know that $S=\frac{n}{2}[2a+(n-1)d]$

So, $S=\frac{2n+1}{2}[2a+(2n+1-1)d]$

$S=(2n+1)(a+nd)$ .......$\left(1\right)$

The alternate terms would be $x_1,x_3,...x_{2n+1}$

So, the number of terms$=n+1$

Here the first term$=a$

and the common diference$=a+2d-a=2d$

$S_1=a+a+2d+a+4d+...+a+2nd$

$=(n+1)a+2\frac{n(n+1)d}{2}$ since there will be $(n+1)$ terms that are $a$ and we will have $2d+4d+...+2nd=2d(1+2+3+...n)$ and sum of $1+2+3+...+n=\frac{n(n+1)}{2}$

$\Rightarrow S_1=(n+1)a+n(n+1)d$

$\Rightarrow S_1=(n+1)(a+nd)$ ......$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get

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