What is the product of n geometric means betw | vedclass

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

Question

(C) Let the $n$ geometric means between $a$ and $b$ be $G_1, G_2, \dots, G_n$.Then $a, G_1, G_2, \dots, G_n, b$ form a Geometric Progression $(GP)$.He

Vedclass · Accepted Answer

$(ab)^{n/2}$

Vedclass · Answer

(C) Let the $n$ geometric means between $a$ and $b$ be $G_1, G_2, \dots, G_n$.Then $a, G_1, G_2, \dots, G_n, b$ form a Geometric Progression $(GP)$.Here,the total number of terms is $n+2$.The last term is $b = a \cdot r^{n+1}$,where $r$ is the common ratio.Thus,$r^{n+1} = \frac{b}{a}$,so $r = (\frac{b}{a})^{\frac{1}{n+1}}$.The product of the $n$ geometric means is $P = G_1 \cdot G_2 \cdot \dots \cdot G_n$.Since $G_k = a \cdot r^k$,we have $P = (a \cdot r^1) \cdot (a \cdot r^2) \dots (a \cdot r^n) = a^n \cdot r^{1+2+\dots+n} = a^n \cdot r^{\frac{n(n+1)}{2}}$.Substituting $r = (\frac{b}{a})^{\frac{1}{n+1}}$,we get $P = a^n \cdot (\frac{b}{a})^{\frac{n(n+1)}{2(n+1)}} = a^n \cdot (\frac{b}{a})^{\frac{n}{2}} = a^n \cdot \frac{b^{n/2}}{a^{n/2}} = a^{n/2} \cdot b^{n/2} = (ab)^{n/2}$.

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

Similar Questions

The sum of the first two terms of a $G.P.$ is $1$ and every term of this series is twice its previous term. Then,the first term will be:

Let ${a_n}$ be the ${n^{th}}$ term of a $G$.$P$. of positive numbers. Let $\sum\limits_{n = 1}^{100} {{a_{2n}}} = \alpha$ and $\sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = \beta$,such that $\alpha \ne \beta$,then the common ratio is

Find the sum of $n$ terms in the geometric progression $\sqrt{7}, \sqrt{21}, 3 \sqrt{7}, \ldots$

If $x, y, z$ are in $G.P.$ and $a^x = b^y = c^z$,then

Suppose the sides of a triangle form a geometric progression with common ratio $r$. Then,$r$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module