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The number of terms of the G.P. $3, \frac{3}{2}, \frac{3}{4}, . . . . . .$ needed to obtain a sum of $\frac{3069}{512}$ is :

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Correct option is B

Here a = 3, r = 1/2.Let the number of terms needed be n, then 3069512=Sn=31−(1/2)n1−1/2⇒ 10231024=1−12n⇒ 12n=11024⇒n=10

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The number of terms of the G.P.3,32,34,......needed to obtain a sum of 3069512 is :

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The number of terms of the G.P. $3, \frac{3}{2}, \frac{3}{4}, . . . . . .$ needed to obtain a sum of $\frac{3069}{512}$ is :