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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the G.P. the sum of whose first two terms is $-4$ and the $5^{th}$ term is $4$ times the $3^{rd}$ term.

$\begin{array}{1 1} 4,-8, 16........\; or - \large\frac{4}{3},-\frac{8}{3}, -\frac{16}{3}...... \\ 4,8, 16........\; or \large\frac{4}{3},\frac{8}{3}, \frac{16}{3}......\\4,-8, 16........\; or - \large\frac{4}{5},-\frac{8}{5}, -\frac{16}{5}...... \\ 4,8, 16........\; or \;\large\frac{4}{5},\frac{8}{5}, \frac{16}{5}......\end{array} $

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1 Answer

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  • $n^{th}$ term of a G.P=$t_n=a.r^{n-1}$
Let the G.P. be $a,ar,ar^2........$
Given that the sum of first $2$ terms of a G.P.$=-4$
and $5^{th}$ term = $4\times 3^{rd}$ term.
$\Rightarrow\:a+ar=-4$ and $t_5=4\times t_3$
$\Rightarrow\:a(1+r)=-4$ and
$a.r^{5-1}=4\times a.r^{3-1}$
$\Rightarrow\:r^4=4r^2$
$\Rightarrow\:r^2=4$ and $r=\pm2$
Substituting the value of $r=2$ we get $a=-\large \frac{4}{3}$
or when $r=-2$, $a=4$
$\therefore\:$ The G.P is $4,-8, 16........$ or $- \large\frac{4}{3},-\frac{8}{3}, -\frac{16}{3}......$
answered Feb 27, 2014 by rvidyagovindarajan_1
 

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