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

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(B) The given numbers are $3^1, 3^2, 3^3, ..., 3^n$.The Geometric Mean $(G.M.)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by $(x_1 \cdot x_2 \cdot

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$3^{\frac{n+1}{2}}$

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(B) The given numbers are $3^1, 3^2, 3^3, ..., 3^n$.The Geometric Mean $(G.M.)$ of $n$ numbers $x_1, x_2, ..., x_n$ is given by $(x_1 \cdot x_2 \cdot ... \cdot x_n)^{1/n}$.$G.M. = (3^1 \cdot 3^2 \cdot 3^3 \cdot ... \cdot 3^n)^{1/n}$Using the property of exponents $a^m \cdot a^n = a^{m+n}$,we get:$G.M. = (3^{1+2+3+...+n})^{1/n}$The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.$G.M. = (3^{\frac{n(n+1)}{2}})^{1/n}$$G.M. = 3^{\frac{n(n+1)}{2n}} = 3^{\frac{n+1}{2}}$.

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

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