If the ratio of the sum of the first three te | vedclass

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

Question

(A) Given the sum of the first $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.We are given $\frac{S_3}{S_6} = \frac{125}{152}$.Substituting

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

(A) Given the sum of the first $n$ terms of a $G.P.$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.We are given $\frac{S_3}{S_6} = \frac{125}{152}$.Substituting the formula: $\frac{a(r^3 - 1)/(r - 1)}{a(r^6 - 1)/(r - 1)} = \frac{125}{152}$.This simplifies to $\frac{r^3 - 1}{r^6 - 1} = \frac{125}{152}$.Since $r^6 - 1 = (r^3 - 1)(r^3 + 1)$,we have $\frac{1}{r^3 + 1} = \frac{125}{152}$.$152 = 125(r^3 + 1) \Rightarrow 152 = 125r^3 + 125$.$125r^3 = 152 - 125 = 27$.$r^3 = \frac{27}{125} = (\frac{3}{5})^3$.Therefore,$r = \frac{3}{5}$.

If the ratio of the sum of the first three terms and the sum of the first six terms of a $G.P.$ is $125 : 152$,then the common ratio $r$ is:

Similar Questions

The $4^{\text{th}}$ term of a $GP$ is $500$ and its common ratio is $\frac{1}{m}$,where $m \in N$. Let $S_n$ denote the sum of the first $n$ terms of this $GP$. If $S_6 > S_5+1$ and $S_7 < S_6+\frac{1}{2}$,then the number of possible values of $m$ is $..........$

Which term of the following sequence: $\sqrt{3}, 3, 3\sqrt{3}, \ldots$ is $729$ (in $^{\text{th}}$)?

The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is

The sum of $100$ terms of the series $0.9 + 0.09 + 0.009 + \dots$ will be

If in a $G.P.$ of $64$ terms,the sum of all the terms is $7$ times the sum of the odd terms of the $G.P.$,then the common ratio of the $G.P.$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module