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5(c)   Which term of the following sequences: 

             \frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,....is \: \: \frac{1}{19683}?

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Given :  GP=\frac{1}{3} , \frac{1}{9} , \frac{1}{27} ,............

a=\frac{1}{3}\, \, \, \, \, and \, \, \, \, \, r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}

n th term is given as  \frac{1}{19683}

a_n=a.r^{n-1}

\Rightarrow \frac{1}{19683}=\frac{1}{3}.(\frac{1}{3})^{n-1}

\Rightarrow \frac{1}{19683}=\frac{1}{3^n}

\Rightarrow \frac{1}{3^9}=\frac{1}{3^n}

\Rightarrow n=9

Thus, n=9.

Posted by

seema garhwal

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