If Sn denote the sum of n terms of an A.P. with first term a and


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) da

(b) d = 2a

(c) a = 2d

(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,

$S_{x}=\frac{x}{2}[2 a+(x-1) d]$'


$\frac{S_{x}}{S_{k x}}=\frac{\frac{x}{2}[2 a+(x-1) d]}{\frac{k x}{2}[2 a+(k x-1) d]}$

$=\frac{[2 a+(x-1) d]}{k[2 a+(k x-1) d]}$

$=\frac{2 a+d x-d}{2 a k+k^{2} x d-k d}$

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,

$\frac{2 a+d x-d}{2 a k+k^{2} x d-k d}=\frac{2 a+d x-2 a}{2 a k+k^{2} x d-2 a k}$

$=\frac{d x}{k^{2} d x}$


Therefore, $2 a=d$

Hence, the correct option is (b).

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