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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the $12^{th}$ term of the $G.P.$ whose $8^{th}$ term is $192$ and common ratio=$2$

$\begin{array}{1 1} 3072 \\ 6144 \\ 1536 \\ 768 \end{array} $

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  • $n^{th}$ tem of a $G.P.=t_n=a.r^{n-1}$ where $1^{st}$ term$ =a$ and common ratio$=r$
Given that the $8^{th}$ term of a G.P.=$192$ and common ratio $r=2$
We know that $t_n=a.r^{n-1}$
$\therefore\:t_8=a.r^{8-1}=a.2^7$
$i.e.,192=a.2^7=128a$
$\Rightarrow\:a=\large\frac{192}{128}=\frac{3}{2}$
$\Rightarrow\:t_{12}=a.r^{12-1}=\large\frac{3}{2}$$.2^{11}=3072$
answered Feb 25, 2014 by rvidyagovindarajan_1
 

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