Question

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

Answer 1

Step: Find the two numbers

Let two numbers be $"a"$ and $"b"$

Since, $14$ is the mean proportional between $a$ and $b$
So, $a,14$ and $b$ will be in continued proportional

$\therefore \frac{a}{14}=\frac{14}{b}$

$\Rightarrow ab=14\times 14$

$\Rightarrow ab=196$

$\Rightarrow a=\frac{196}{b}$ ....(i)

Now, $112$ is the third proportion of $a$ and $b$

$\therefore \frac{a}{b}=\frac{b}{112}$

$\Rightarrow b^2=112a$ [By cross product]

$\Rightarrow b^2=112\times \frac{196}{b}$ [Form (i) ]

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

$\Rightarrow b^3=21952$

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

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

$\therefore b=\sqrt[3]{3{28}^3}$

$\therefore b=28$

From Equation (i)

$a=\frac{196}{28}=7$

Hence, required two numbers are $7$ and $28$.