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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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The $4^{th}$ term of a G.P. is square of its second term, and the $1^{st}$ term is $-3$. Find its $7^{th}$ term.

$\begin{array}{1 1}-2187 \\ 2187 \\ -1183 \\ -2347 \end{array} $

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  • $n^{th}\:term$ of a G.P.$=a.r^{n-1}$ where $1^{st}\:term=a$ and $common\: ratio=r$
Given: In a G.P. $4^{th}\:term=(2^{nd}\:term)^2\:\:and 1^{st}\:term=-3$
$i.e., \:t_4=(t_2)^2\:\;and\:\:a=-3$
We know that $t_n=a.r^{n-1}$
$\therefore\:a.r^{4-1}=(a.r^{2-1})^2$
$\Rightarrow\:a.r^3=a^2.r^2$
$r=a$
Substituting the value of $a=r=-3$ we get
$t_7=a.r^{7-1}=a.r^6=-3.(-3)^6=-3^7=-2187$
answered Feb 25, 2014 by rvidyagovindarajan_1
 

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