The product of three geometric means between | vedclass

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(d) We have $4,\;{g_1},\;{g_2},\;{g_3},\;\frac{1}{4}$ is a $G.P.$

Here $a = 4$,

${g_1} = ar = 4 	imes r,\;\;\;{g_2} = a{r^2},\;\;\;{g_3} = a{r^

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(d) We have $4,\;{g_1},\;{g_2},\;{g_3},\;\frac{1}{4}$ is a $G.P.$

Here $a = 4$,

${g_1} = ar = 4 	imes r,\;\;\;{g_2} = a{r^2},\;\;\;{g_3} = a{r^3}$

and ${g_4} = a{r^4} = 4 	imes {r^4} = \frac{1}{4}$

$ \Rightarrow {r^4} = \frac{1}{{16}} = {\left( {\frac{1}{2}} ight)^4}$

$\Rightarrow r = \frac{1}{2}$

Now product of three $G.M.$ ${g_1}.\;{g_2}.\;{g_3} = ar.\;a{r^2}.\;a{r^3}$

$ = {a^3}{r^6} = {4^3} 	imes {\left( {\frac{1}{2}} ight)^6} = \frac{{{4^3}}}{{{4^3}}} = 1$.

Note : The product of &lsquo;$n$&rsquo; geometric means between &lsquo;$a$&rsquo; and $\frac{1}{a}$ is always equal to $1$.

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

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