If $S_{n}$ denote the sum of $n$ terms of an A.P. with first term a and common difference $d$ such that $\frac{S x}{S k x}$ is independent of $x$, then
(a) d= a
(b) d = 2a
(c) a = 2d
(d) d = −a
Here, we are given an A.P. with a as the first term and d as the common difference. The sum of n terms of the A.P. is given by Sn.
We need to find the relation between $a$ and $d$ such that $\frac{S_{x}}{S_{k 1}}$ is independent of
So, let us first find the values of Sx and Skx using the following formula for the sum of n terms of an A.P.,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Where; $\alpha=$ first term for the given A.P.
$d=$ common difference of the given A.P.
$n=$ number of terms
So, we get,
$S_{x}=\frac{x}{2}[2 a+(x-1) d]$'
So,
$\frac{S_{x}}{S_{k x}}=\frac{\frac{x}{2}[2 a+(x-1) d]}{\frac{k x}{2}[2 a+(k x-1) d]}$
$=\frac{[2 a+(x-1) d]}{k[2 a+(k x-1) d]}$
$=\frac{2 a+d x-d}{2 a k+k^{2} x d-k d}$
Now, to get a term independent of x we have to eliminate the other terms, so we get
$2 a-d=0$
$2 a=d$
So, if we substitute $2 a=d$, we get,
$\frac{2 a+d x-d}{2 a k+k^{2} x d-k d}=\frac{2 a+d x-2 a}{2 a k+k^{2} x d-2 a k}$
$=\frac{d x}{k^{2} d x}$
$=\frac{1}{k^{2}}$
Therefore, $2 a=d$
Hence, the correct option is (b).
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