Question

If in an A.P., $S_1=T_1+T_2+T_3+...+T_n$ $[n$ is odd$]$ and $S_2=T_1+T_3+T_5+...+T_n$ 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 $\frac{2n}{n+1}$

If there are 9 odd terms then $T_2+T_4+T_6+T_8$ will be $\frac{9-1}{2}=4$ in number and
$T_1+T_3+T_5+T_7+T_9$ will be $\frac{9+1}{2}=5$ in number
Hence $S_1$ is an A.P of $n$ terms but $S_2$ is an A.P of $\frac{n+1}{2}$ terms with common difference of $2d$
$S_1=\frac{n}{2}[2a+(n-1)d]$ ...(1)
$S_2=\frac{1}{2}\left[\frac{n+1}{2}\right][2a+(\frac{n+1}{2}-1)2d]=\frac{n+1}{4}[2a+(n-1)d]$ ...(2)
From (1) and (2)
$\frac{S_1}{S_2}=\frac{2n}{n+1}$