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In a GP `a_1=3,a_n=96` and `S_n=189` .Find the number of terms |
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Answer» Here, `a=3, l=96` and `S_(n)=189`. Let the common ratio of the given GP be r. Then, `S_(n)=((lr-a))/((r-1)) rArr ((96r-3))/((r-1))=189` `rArr (96r-3)=(189r-189)` `rArr 93r=186 rArr r=2`. Now, `l=ar^(n-1) rArr 3xx2^(n-1)=96` `rArr 2^(n-1)=32=2^(5) rArr n-1=5 rArr n=6`. Hence, `n=6`. |
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