If n geometric means between a and b are G_1, | vedclass

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(C) Given that $G$ is the geometric mean between $a$ and $b$,we have $G = (ab)^{1/2}$.If $G_1, G_2, ..., G_n$ are $n$ geometric means between $a$ and

Vedclass · Accepted Answer

$G_1 \cdot G_2 \cdot ... \cdot G_n = G^n$

Vedclass · Answer

(C) Given that $G$ is the geometric mean between $a$ and $b$,we have $G = (ab)^{1/2}$.If $G_1, G_2, ..., G_n$ are $n$ geometric means between $a$ and $b$,then $a, G_1, G_2, ..., G_n, b$ form a geometric progression with $n+2$ terms.Let $r$ be the common ratio. Then $b = a r^{n+1}$,which implies $r = (b/a)^{1/(n+1)}$.The product of the $n$ geometric means is $P = G_1 \cdot G_2 \cdot ... \cdot G_n = (ar) \cdot (ar^2) \cdot ... \cdot (ar^n) = a^n r^{1+2+...+n} = a^n r^{n(n+1)/2}$.Substituting $r = (b/a)^{1/(n+1)}$,we get:$P = a^n \left( \frac{b}{a} \right)^{\frac{n(n+1)}{2(n+1)}} = a^n \left( \frac{b}{a} \right)^{n/2} = a^n \cdot \frac{b^{n/2}}{a^{n/2}} = a^{n/2} b^{n/2} = (ab)^{n/2}$.Since $G = (ab)^{1/2}$,it follows that $G^n = ((ab)^{1/2})^n = (ab)^{n/2}$.Therefore,$G_1 \cdot G_2 \cdot ... \cdot G_n = G^n$.

If $n$ geometric means between $a$ and $b$ are $G_1, G_2, ..., G_n$ and a single geometric mean is $G$,then the true relation is:

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