STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Here $a = 4$,

${g_1} = ar = 4 \times r,\;\;\;{g_2} = a{r^2},\;\;\;{g_3} = a{r^3}$

and ${g_4} = a{r^4} = 4 \times {r^4} = \frac{1}{4}$

$ \Rightarrow {r^4} = \frac{1}{{16}} = {\left( {\frac{1}{2}} \right)^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} \times {\left( {\frac{1}{2}} \right)^6} = \frac{{{4^3}}}{{{4^3}}} = 1$.

Note : The product of ‘$n$’ geometric means between ‘$a$’ and $\frac{1}{a}$ is always equal to $1$.