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In a GP `a_1=3,a_n=96` and `S_n=189` .Find the number of terms

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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