1.

The sum of some terms of G. P.is 315 whose first term and the common ratio are 5 and 2, respectively. Findthe last term and the number of terms.

Answer» Let the given GP contain n terms, Then,
`a=5, r=2 gt 1` and `S_(n)=315`.
`:. S_(n)=315 rArr (a(r^(n)-1))/((r-1))=315`
`rArr (5xx(2^(n)-1))/((2-1))=315`
`rArr (2^(n)-1)=63 rArr 2^(n) =64=2^(6) rArr n=6`.
Last term `=ar^((n-1))=5xx2^((6-1))=((5xx2^(5))=(5xx32))=160`.
Hence, the given GP contains 6 terms and its last term is 160.


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