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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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.

Similar Questions

How many terms of the $G.P.$ $3, \frac{3}{2}, \frac{3}{4}, \ldots$ are needed to give the sum $\frac{3069}{512} ?$

Let $a, b, c, d$ and $p$ be any non-zero distinct real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2} - 2(ab+bc+cd) p + (b^{2}+c^{2}+d^{2}) = 0$. Then:

The sum of infinite terms of a $G.P.$ is $x$ and on squaring each term of it,the sum becomes $y$. Then the common ratio of this series is:

The $G.M.$ of the numbers $3, 3^2, 3^3, ..., 3^n$ is

If the sum of an infinite $GP$ is $9$ and the sum of the first two terms is $5$,then their common ratio is .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module