In a geometric progression,if the ratio of th | vedclass

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

Question

(C) Let the geometric progression be $a, ar, ar^2, ar^3, ar^4$.The sum of the first $5$ terms is $S_5 = a + ar + ar^2 + ar^3 + ar^4 = \frac{a(r^5 - 1)

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(C) Let the geometric progression be $a, ar, ar^2, ar^3, ar^4$.The sum of the first $5$ terms is $S_5 = a + ar + ar^2 + ar^3 + ar^4 = \frac{a(r^5 - 1)}{r - 1}$.The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$.The sum of these reciprocals is $S'_5 = \frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4} = \frac{1}{a} \left( \frac{1 - (1/r)^5}{1 - 1/r} \right) = \frac{1}{a} \left( \frac{(r^5 - 1)/r^5}{(r - 1)/r} \right) = \frac{1}{a} \cdot \frac{r^5 - 1}{r^5} \cdot \frac{r}{r - 1} = \frac{r^5 - 1}{a r^4 (r - 1)}$.Given the ratio $\frac{S_5}{S'_5} = 49$:$\frac{\frac{a(r^5 - 1)}{r - 1}}{\frac{r^5 - 1}{a r^4 (r - 1)}} = 49$ $\Rightarrow a^2 r^4 = 49$ $\Rightarrow ar^2 = 7$ (since $a$ and $r$ are real).Given the sum of the first and third term is $35$:$a + ar^2 = 35$.Substituting $ar^2 = 7$ into the equation:$a + 7 = 35 \Rightarrow a = 28$.

In a geometric progression,if the ratio of the sum of the first $5$ terms to the sum of their reciprocals is $49$,and the sum of the first and the third term is $35$,then the first term of this geometric progression is:

Similar Questions

If $a_1, a_2, \dots, a_{50}$ are in a geometric progression,then $\frac{a_1 - a_3 + a_5 - \dots + a_{49}}{a_2 - a_4 + a_6 - \dots + a_{50}} = \dots$

There are two such pairs of non-zero real values of $a$ and $b$,i.e.,$(a_1, b_1)$ and $(a_2, b_2)$,for which $2a+b, a-b, a+3b$ are three consecutive terms of a $G.P.$. Then the value of $2(a_1b_2 + a_2b_1) + 9a_1a_2$ is-

What is the product of $n$ geometric means between $a$ and $b$?

Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a $G$.$P$. with common ratio $r = \frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \ldots + a_{10} = 62$, then $a_1$ is equal to:

If the $p^{\text{th}}$,$q^{\text{th}}$,and $r^{\text{th}}$ terms of a $G.P.$ are $a$,$b$,and $c$ respectively,prove that $a^{q-r} b^{r-p} c^{p-q} = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module