Question

Find two numbers such that the mean proportional between them is $14$ and the third proportional to them is $112$

Answer 1

Let the required number be $x\&y$.

$\therefore 14$is the mean proportional between $x\&y$.

Then,

$\frac{x}{14}=\frac{14}{y}$

$\Rightarrow xy=14\times 14=196$

$\Rightarrow x=\frac{196}{y}.......\left(1\right)$

And given that, $112$ is the third proportional to $x\&y$.

Then, $x:y::y:112$

$i.e.\frac{x}{y}=\frac{y}{112}$

$\Rightarrow y^2=112x.......\left(2\right)$

Putting the value of $x$ by equation (1) to (2) and we get,

$\Rightarrow y^2=112\times \frac{196}{y}$

$\Rightarrow y^3=112\times 196$

$\Rightarrow y^3=2\times 2\times 2\times 2\times 7\times 2\times 7\times 2\times 7$

$\Rightarrow y^3=28\times 28\times 28$

$\Rightarrow y^3={28}^3$

$\Rightarrow y=28$

Put the value of y in equation (1), and we get

$x=\frac{196}{28}$

$x=7$

Hence, this is the answer.