Question

The mean proportional between $9$ and $16$ is
$\left(a\right)12.5$
$\left(b\right)12$
$\left(c\right)5$
$\left(d\right)$ none of these

Answer 1

Given numbers are $9$ and $16$.

We know that, if $a:b::c:d$ then $ad=bc$

i.e., Product of extremes $=$ Product of the means.

Suppose that the mean proportional is $x$

Then, $9:x::x:16$

$\Rightarrow 9\times 16=x\times x$

$\Rightarrow 144=x^2$

$\Rightarrow \sqrt{144}=x$

$\Rightarrow 12=x$

Hence, Option $\left(c\right)$ is correct.

(or)

Note: The mean proportional between $a$ and $b$ is $\sqrt{ab}$.

$\Rightarrow$ The mean proportional between $9$ and $16$ is

$=\sqrt{16\times 9}=\sqrt{144}=12$