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

Vedclass · Accepted Answer

$\frac{3}{5}$

Vedclass · Answer

(A) The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n - 1)}{r - 1}$.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 $\frac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \frac{125}{152}$.$\frac{1}{r^3 + 1} = \frac{125}{152}$.$r^3 + 1 = \frac{152}{125}$.$r^3 = \frac{152}{125} - 1 = \frac{152 - 125}{125} = \frac{27}{125}$.Taking the cube root on both sides,$r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5}$.

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 mean of the series $a, a + nd, a + 2nd$ is

If $a_1, a_2, a_3, \dots$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is

If the sum of an infinite $G.P.$ and the sum of the squares of its terms is $3$,then the common ratio of the first series is

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

The geometric and harmonic means of two numbers $x_1$ and $x_2$ are $18$ and $16\frac{8}{13}$ respectively. The value of $|x_1 - x_2|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module