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 $S_5 = a(1+r+r^2+r^3+r^4) = a\frac{r^5-1}{r-1}$.The r

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 $S_5 = a(1+r+r^2+r^3+r^4) = a\frac{r^5-1}{r-1}$.The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}, \frac{1}{ar^4}$.The sum of the reciprocals is $S'_5 = \frac{1}{a}(1 + \frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4}) = \frac{1}{a} \frac{\frac{1}{r^5}-1}{\frac{1}{r}-1} = \frac{1}{a} \frac{1-r^5}{r^4(1-r)} = \frac{r^5-1}{ar^4(r-1)}$.Given the ratio $\frac{S_5}{S'_5} = 49$:$\frac{a(r^5-1)/(r-1)}{(r^5-1)/(ar^4(r-1))} = 49$$a^2 r^4 = 49$$(ar^2)^2 = 7^2 \Rightarrow ar^2 = 7$.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$$a = 28$.

In a geometric progression,if the ratio of the sum of 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

Determine $k,$ so that $k+2, 4k-6$ and $3k-2$ are three consecutive terms of an $A.P.$

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

If $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$,then $n =$

If the sum of the series $1 + \frac{2}{x} + \frac{4}{x^2} + \frac{8}{x^3} + \dots \infty$ is a finite number,then

The sum to infinity of a geometric progression is $4/3$ and the first term is $3/4$. The common ratio is

Vedclass Test Series

Exam Paper Generator

Online Exam Module