The G.M. of the numbers 3, 3^2, 3^3, ..., 3^n | vedclass

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(D) The $G.M.$ of $n$ numbers $a_1, a_2, ..., a_n$ is given by $(a_1 	imes a_2 	imes ... 	imes a_n)^{1/n}$.Here,the numbers are $3^1, 3^2, 3^3, ...

Vedclass · Accepted Answer

$3^{(n + 1)/2}$

Vedclass · Answer

(D) The $G.M.$ of $n$ numbers $a_1, a_2, ..., a_n$ is given by $(a_1 	imes a_2 	imes ... 	imes a_n)^{1/n}$.Here,the numbers are $3^1, 3^2, 3^3, ..., 3^n$.$G.M. = (3^1 	imes 3^2 	imes 3^3 	imes ... 	imes 3^n)^{1/n}$$G.M. = (3^{1 + 2 + 3 + ... + n})^{1/n}$Using the sum of the first $n$ natural numbers formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:$G.M. = (3^{\frac{n(n+1)}{2}})^{1/n}$$G.M. = 3^{\frac{n(n+1)}{2n}}$$G.M. = 3^{\frac{n+1}{2}}$

The $G.M.$ of the numbers $3, 3^2, 3^3, ..., 3^n$ is

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