If n geometric means are inserted between a a | vedclass

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

Question

(C) If $n$ geometric means $g_1, g_2, \dots, g_n$ are inserted between two positive real numbers $a$ and $b$,then the sequence $a, g_1, g_2, \dots, g_

Vedclass · Accepted Answer

$a\left( \frac{b}{a} \right)^{\frac{n}{n+1}}$

Vedclass · Answer

(C) If $n$ geometric means $g_1, g_2, \dots, g_n$ are inserted between two positive real numbers $a$ and $b$,then the sequence $a, g_1, g_2, \dots, g_n, b$ forms a Geometric Progression $(G.P.)$.Let $r$ be the common ratio of this $G.P.$ The total number of terms in this sequence is $n+2$.Thus,the last term $b$ can be expressed as $b = a \cdot r^{(n+2)-1} = a \cdot r^{n+1}$.Solving for $r$,we get $r = \left( \frac{b}{a} \right)^{\frac{1}{n+1}}$.The $n^{th}$ geometric mean is $g_n = a \cdot r^n$.Substituting the value of $r$,we get $g_n = a \left( \left( \frac{b}{a} \right)^{\frac{1}{n+1}} \right)^n = a \left( \frac{b}{a} \right)^{\frac{n}{n+1}}$.

If $n$ geometric means are inserted between $a$ and $b$,then the $n^{th}$ geometric mean will be:

Similar Questions

If every term of a $G.P.$ with positive terms is the sum of its two previous terms,then the common ratio of the series is

If the first term of an $A.P.$ is $10$,the last term is $50$,and the sum of all the terms is $300$,then the number of terms is:

If $3 + \frac{1}{4} (3 + d) + \frac{1}{4^2} (3 + 2d) + \dots \infty = 8$,then the value of $d$ is:

Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ terms respectively of a non-constant $A.P.$ If these are also the three consecutive terms of a $G.P.$,then $\frac{a}{c}$ is equal to

If $1 + \cos \alpha + \cos^2 \alpha + \dots \infty = 2 - \sqrt{2}$,then $\alpha$ $(0 < \alpha < \pi)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module