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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Find the $20^{th}$ term of the $G.P.\:\:\large\frac{5}{2},\frac{5}{4},\frac{5}{8},..........$

$\begin{array}{1 1}\large\frac{5}{2}.\big(\frac{1}{2}\big)^{19} \\\large\frac{5}{2}.\big(\frac{1}{2}\big)^{20} \\\large\frac{5}{4}.\big(\frac{1}{2}\big)^{19} \\ \large\frac{5}{4}.\big(\frac{1}{2}\big)^{20} \end{array} $

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  • $n^{th}\:term$ of a $G.P.=t_n=a.r^{n-1}$ where $r$ is common ratio and $a=first\:term$
Given G.P. is $\large\frac{5}{2},\frac{5}{4},\frac{5}{8}........$
$\Rightarrow\:First\: term=a=\large\frac{5}{2},\:\: $$ common\: ratio=r=\large\frac{5/4}{5/2}=\frac{1}{2}$
We know that $t_n=a.r^{n-1}=\large\frac{5}{2}.\big(\frac{1}{2}\big)^{n-1}$
$\therefore\:t_{20}=\large\frac{5}{2}.\big(\frac{1}{2}\big)^{19}$
answered Feb 25, 2014 by rvidyagovindarajan_1
 

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