Question

In an A.P. if
$S_1=T_1+T_2+T_3+$______$+T_n$(n odd)
$S_2=T_1+T_3+T_5+$______$+T_n$, then $S_1/S_2=$.

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