If n geometric means between a and b be {G_1} | vedclass

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Question

(c) Here $G = {(ab)^{1/2}}$ and

${G_1} = a{r^1},\;{G_2} = a{r^2},........{G_n} = a{r^n}$

Therefore ${G_1}.\;{G_2}.\;{G_3}.....{G_n} = {a^n}{r^{1

Vedclass · Accepted Answer

${G_1}.{G_2}........{G_n} = {G^n}$

Vedclass · Answer

(c) Here $G = {(ab)^{1/2}}$ and

${G_1} = a{r^1},\;{G_2} = a{r^2},........{G_n} = a{r^n}$

Therefore ${G_1}.\;{G_2}.\;{G_3}.....{G_n} = {a^n}{r^{1 + 2 + ... + n}} = {a^n}{r^{n(n + 1)/2}}$

But $a{r^{n + 1}} = b$

$\Rightarrow r = {\left( {\frac{b}{a}} \right)^{1/(n + 1)}}$

Therefore, the required product is

${a^n}{\left( {\frac{b}{a}} \right)^{1/(n + 1)\;.\;n(n + 1)/2}} = {(ab)^{n/2}} $

$= {\left\{ {{{(ab)}^{1/2}}} \right\}^n} = {G^n}$.

If $n$ geometric means between $a$ and $b$ be ${G_1},\;{G_2},\;.....$${G_n}$ and a geometric mean be $G$, then the true relation is

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