If a_1, a_2, a_3, ldots, a_{10} is a geometri | vedclass

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

Question

(B) Given that $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression $(GP)$ with first term $a$ and common ratio $r$. We are given $\frac{a_3}{a_

Vedclass · Accepted Answer

$5^4$

Vedclass · Answer

(B) Given that $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression $(GP)$ with first term $a$ and common ratio $r$. We are given $\frac{a_3}{a_1} = 25$. Since $a_n = ar^{n-1}$,we have $a_3 = ar^2$ and $a_1 = a$. Thus,$\frac{ar^2}{a} = 25$,which implies $r^2 = 25$. We need to find the value of $\frac{a_9}{a_5}$. Using the formula for the $n$-th term,$\frac{a_9}{a_5} = \frac{ar^8}{ar^4} = r^4$. Since $r^2 = 25$,we have $r^4 = (r^2)^2 = (25)^2 = (5^2)^2 = 5^4$.

If $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression and $\frac{a_3}{a_1}=25$,then $\frac{a_9}{a_5}$ equals

Similar Questions

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 $7^{th}$ term of the sequence $\sqrt{2}, \sqrt{10}, 5\sqrt{2}, \dots$ is

Find the $20^{\text{th}}$ and $n^{\text{th}}$ terms of the $G.P.$ $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$

Find the sum to the indicated number of terms in the geometric progression: $1, -a, a^{2}, -a^{3}, \ldots$ to $n$ terms (if $a \neq -1$).

If $a^2 + b^2 + 16c^2 = 2(3ab + 6bc + 4ac)$,where $a, b, c$ are non-zero numbers,then $a, b, c$ are in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module