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8. Sequences and Series
easy
The product of three geometric means between $4$ and $\frac{1}{4}$ will be
A
$4$
B
$2$
C
$- 1$
D
$1$
Solution
(d) We have $4,\;{g_1},\;{g_2},\;{g_3},\;\frac{1}{4}$ is a $G.P.$
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 $1$.
Standard 11
Mathematics
