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8. Sequences and Series
easy
If three geometric means be inserted between $2$ and $32$, then the third geometric mean will be
A
$8$
B
$4$
C
$16$
D
$12$
Solution
(c) $2,\,{g_1},\;{g_2},\;{g_3},\;32$ where
$a = 2,\;ar = {g_1},\;a{r^2} = {g_2},\;a{r^3} = {g_3}$ and $a{r^4} = 32$
Now $2 \times {r^4} = 32$
$\Rightarrow {r^4} = 16 = {(2)^4} $
$\Rightarrow r = 2$.
Then third geometric mean $ = a{r^3} = 2 \times {2^3} = 16$.
Aliter : By formula, ${G_3} = 2.{\left( {\frac{{32}}{2}} \right)^{3/4}} = 2\;.\;8 = 16$.
Standard 11
Mathematics