The sum of some terms of a G.P. is 315,whose | vedclass

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(A) Let the sum of $n$ terms of the $G.P.$ be $S_n = 315$.The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.Given t

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Last term: $160$,Number of terms: $6$

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(A) Let the sum of $n$ terms of the $G.P.$ be $S_n = 315$.The formula for the sum of $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.Given that the first term $a = 5$ and the common ratio $r = 2$.Substituting the values into the formula: $315 = \frac{5(2^n - 1)}{2 - 1}$.$315 = 5(2^n - 1)$.$2^n - 1 = \frac{315}{5} = 63$.$2^n = 64 = 2^6$.Therefore,$n = 6$.The last term of the $G.P.$ is the $n^{th}$ term,given by $a_n = a \cdot r^{n-1}$.$a_6 = 5 \cdot 2^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160$.Thus,the last term is $160$ and the number of terms is $6$.

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

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