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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Which term of the sequence $2,2\sqrt 2,4,.......\:\:is\:\:128?$

$\begin{array}{1 1}13 th \\ 11th \\ 12th \\ 14 th \end{array} $

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  • $n^{th}\:term$ of a G.P.$=t_n=a.r^{n-1}$ where $a=1^{st}\:term\:and\:d=common \:ratio.$
Given series is $2,2\sqrt 2,4,.........$
This series is G.P. since the ratio of any two successive terms is same.
first term$=a=2$ and common ratio $r=\large\frac{2\sqrt 2}{2}$$=\sqrt 2$
We know that $n^{th}$ term $=t_n=a.r^{n-1}$
Given that $t_n=128$
$\Rightarrow\:128=2.(\sqrt 2)^{n-1}$
$\sqrt 2=2^{1/2}$ and $128=2^7$
Since the base on either side is 2, power on both sides are equal.
or $n=13$
$i.e., \:\:13^{th}$ term of the sequence is $128$
answered Feb 25, 2014 by rvidyagovindarajan_1

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