Question

In an A.P., let $T_n$ denotes the $n^{th}$ term and $S_n$ denotes the sum of first $n$ terms. If $T_7=\frac{1}{9}$ and $T_9=\frac{1}{7},$ then the value of $S_{63}$ is

• A. $32$
• B. $49$
• C. $56$
• D. $63$

Answer 1

The correct option is A $32$

Let $a$ and $d$ be the first term and common difference of A.P. respectively.
Then, $T_7=a+6d=\frac{1}{9}\cdots \left(1\right)$
and $T_9=a+8d=\frac{1}{7}\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$a=\frac{1}{63},d=\frac{1}{63}\therefore S_{63}=\frac{63}{2}[2a+(63-1\left)d\right]=\frac{63}{2}[2\times \frac{1}{63}+62\times \frac{1}{63}]=32$