Question

Consider an AP with first term $a$ and the common difference $d$. Let $S_k$ denote the sum of its first $k$ terms, if $\frac{S_{kx}}{S_x}$ is independent of $x$ then

• A. $a=\frac{d}{2}$
• B. $a=d$
• C. $a=2d$
• D. none of these

Answer 1

The correct option is A $a=\frac{d}{2}$

Given, first term =$a$, common difference $=d$

By using $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ we have,

$\therefore \frac{Skx}{Sx}=\frac{\frac{kx}{2}(2a+(kx-1\left)d\right)}{\frac{x}{2}(2a+(x+1\left)d\right)}$

$\Rightarrow \frac{Skx}{Sx}=\frac{k(2a+(kx-1\left)d\right)}{(2a+(x+1\left)d\right)}$

$\Rightarrow \frac{Skx}{Sx}=\frac{k(2a-d+kxd)}{2a+xd-d}$

Now this will be independent of $x$ only when $2a=d$ or $a=\frac{d}{2}$