Question

If $S_1$ is the sum of an arithmetic progression of ${}^{\prime }{n}^{\prime }$ odd number of terms and $S_2$ the sum of the terms of the series in odd places, then $\frac{S_1}{S_2}$ =

• A. $\frac{2n}{n+1}$
• B. $\frac{n}{n+1}$
• C. $\frac{n+1}{2n}$
• D. $\frac{n+1}{n}$

Answer 1

The correct option is A $\frac{2n}{n+1}$

Let $n=2m+1$

Then, $S_1=\frac{n}{2}[2a_1+(n-1)d]=\frac{2m+1}{2}[2a_1+2md]$

$=(2m+1)(a_1+md)=n(a+\frac{n-1}{2}d)$

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

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

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

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