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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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Which term of the sequence $\large\frac{1}{3},\frac{1}{9},\frac{1}{27}.......$ is $\large\frac{1}{19683}$?

$\begin{array}{1 1}7th\;term \\ 8th\;term \\ 9th \;term \\ 10th\;term \end{array} $

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1 Answer

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  • $n^{th}$ term of a G.P$=t_n=a.r^{n-1}$ where $a=1^{st}\:term\:and\:r=common\:ratio.$
The given sequence is $\large\frac{1}{3},\frac{1}{9},\frac{1}{27}..........$
This sequence is G.P. since the ratio of any two successive terms is same.
In this sequence $1^{st}$ term $a=\large\frac{1}{3}$
common ratio=$r=\large\frac{1/9}{1/3}=\large\frac{1}{3}$
We know that $t_n=a.r^{n-1}$
Given that $t_n=\large\frac{1}{19683}$
$\Rightarrow\:\large\frac{1}{19683}=\frac{1}{3}.(\frac{1}{3})^{n-1}$
$19683=3^9$
$\Rightarrow\:\large\frac{1}{19683}=(\frac{1}{3})^n$
$\Rightarrow\:\large\frac{1}{3^9}=\frac{1}{3^n}$
$\therefore\:n=9$
$i.e.,\:\:9^{th}$ term of the given sequence is $\large\frac{1}{19683}$
answered Feb 25, 2014 by rvidyagovindarajan_1
 

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