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Updated on : 2022-09-05

Solution

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From the question, it is given that,

Sum of the terms $S_{n}=3069/512$

First term $a=3$

Common ratio $r=(3/2)/3$

$=(3/2)×(1/3)$

$=1/2$

We know that,

$S_{n}=a(1−r_{n})/(1−r)$

$(3069/512)=3[1−(1/2)_{n}]/(1−1/2)$

$(3069/512)=(2×3)[1−(1/2)_{n}]$

$1−(1/2)_{n}=3069/(512×6)$

$1−(1/2)_{n}=1023/1024$

$(1/2)_{n}=1−(1023/1024)$

$(1/2)_{n}=(1024−1023)/1024$

$(1/2)_{n}=1/1024$

$(1/2)_{n}=(1/2)_{10}$

By comparing both LHS and RHS, $n=10$

Therefore, there are 10 terms are in the G.P.

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