Question

Let $T_1,T_2,T_3,\dots$ be terms of an A.P. If $S_1=T_1+T_2+T_3+\cdots +T_n$ and $S_2=T_2+T_4+T_6+\cdots +T_{n-1}$, where $n$ is odd, then the value of $\frac{S_1}{S_2}$ is

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

Answer 1

The correct option is D $\frac{2n}{n-1}$

Let $d$ be the common difference of A.P.
$S_1$ contains $n$ terms with common difference $d$.
But $S_2$ contains $\frac{n-1}{2}$ terms with common difference $2d$.

$\therefore S_1=\frac{n}{2}(T_1+T_n)$
and $S_2=\frac{\frac{n-1}{2}}{2}(T_2+T_{n-1})$
$=\frac{\frac{n-1}{2}}{2}(T_1+T_n)$

We know that, in a finite A.P., the sum of the terms equidistant from the beginning and the end is equal.
So, $T_1+T_n=T_2+T_{n-1}$
Hence, $\frac{S_1}{S_2}=\frac{2n}{n-1}$