$(a)\;a=d\qquad(b)\;a=\large\frac{d}{2}\qquad(c)\;a=2d\qquad(d)\;None\;of\;these$

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Answer : (b) $\;a=\large\frac{d}{2}$

Explanation : $\;\large\frac{S_{nx}}{S_{x}}=\large\frac{\large\frac{nx}{2}\;[2a+(nx-1)d]}{\large\frac{x}{2}\;[2a+(x-1)d]}=k\;(say)$

$2an+(nx-1)dn=k\;[2a+(n-1)d]$

$2an+n^2dx-nd=2ak+kdx-kd$

$2an-2ak-nd+kd+(n^{2}d-kd)x=0$

Since above term is independent of x

$n^2d-kd=0$

$k=n^2$

$2an-2ak-nd+kd=0$

$2an-2an^2-nd+n^2d=0$

$2an(1-n)-nd(1-n)=0$

$(1-n)n(2a-d)=0$

$2a-d=0$

$a=\large\frac{d}{2}\;.$

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