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13.   How many terms of G.P. 3 , 3 ^ 2 , 3 ^ 3, … are needed to give the sum 120? 

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G.P.=    3 , 3 ^ 2 , 3 ^ 3, …............

Sum =120

These terms are GP with a=3 and r=3.

S_n=\frac{a(1-r^n)}{1-r}

120=\frac{3(1-3^n)}{1-3}

120\times \frac{-2}{3}=(1-3^n)

-80=(1-3^n)

\Rightarrow 3^n=1+80=81

\Rightarrow 3^n=81

\Rightarrow 3^n=3^4

\Rightarrow n=4

Hence, we have value of n as 4 to get sum of 120.

Posted by

seema garhwal

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