MCQ
If three geometric means be inserted between $2$ and $32$, then the third geometric mean will be
- A$8$
- B$4$
- ✓$16$
- D$12$
$a = 2,\;ar = {g_1},\;a{r^2} = {g_2},\;a{r^3} = {g_3}$ 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, ${G_3} = 2.{\left( {\frac{{32}}{2}} \right)^{3/4}} = 2\;.\;8 = 16$.
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