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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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If $a, b, c$----$l$ is a GP, find the sum of the series.

$\begin{array}{1 1}l-(b/a)^n \\ \large\frac{(b/a)^{n-1}+l}{2} \\ abl \\ \large\frac{bl-a^2}{b-a} \end{array}$

Can you answer this question?
 
 

1 Answer

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Answer : (d) $\large\frac{bl-a^2}{b-a}$
$Explanation :\;r=\large\frac{b}{a}\quad\;l=a\;(b/a)^{n-1} $
$\large\frac{b}{a}\;l=a\;(b/a)^{n-1}\;\large\frac{b}{a}=a\;(b/a)^n$
$r^n=(b/a)^n=\large\frac{bl}{a^2}$
$S=\large\frac{a(r^n-1)}{r-1}=\large\frac{a(\large\frac{bl}{a^2}-1)}{(b/a)-1}$
$S=\large\frac{bl-a^2}{b-a}.$
answered Jan 3, 2014 by yamini.v
edited Jan 24, 2014
 

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