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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. |
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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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