Question

If, $S_1$ is the sum of an arithmetic progression of 'n' 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 we have a series of odd no.s

$a,a+d,aa+2d+....,a+(n-1)d$ where $n$ is odd

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

As $S_2$ is the sum of terms of the series in odd place

So,

$S_2=a+a+2d+a+4d+......+(a+(\frac{n-1}{2}\left)d\right)$

So total no. of terms are $\frac{n+1}{2}$ and difference $2d$

$S_2=\frac{n+1}{4}[2a+(\frac{n+1}{2}-1\left)2d\right]$

$\frac{S_1}{S_2}=\frac{n/2(2a+(n-1\left)d\right))}{\left[\right(n+1\left)/4\right](2a+(n-1\left)d\right)}=\frac{2n}{n+1}$