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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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For an AP with first term $a$ and common difference $d$, if $\large\frac{S_{nx}}{S_{x}}\:(S_{r}$ denotes sum upto $r$ term$)$ is independent of $x$ then,

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

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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
$n^2d-kd=0$
$k=n^2$
$2an-2ak-nd+kd=0$
$2an-2an^2-nd+n^2d=0$
$2an(1-n)-nd(1-n)=0$
$(1-n)n(2a-d)=0$
$2a-d=0$
$a=\large\frac{d}{2}\;.$
answered Jan 23, 2014 by yamini.v
 

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