Find two numbers such that the mean proportion between them is $14$ and the third proportion to them is $112$.
Let the two numbers be $p$ and $q$.
Given: Mean proportion between $p$ and $q$ is $14$.
$\implies$ $p$, $14$ and $q$ are in continued proportion.
i.e. $\;$ $p : 14 = 14 : q$
i.e. $\;$ $14^2 = 196 = p \times q$ $\;\;\; \cdots \; (1)$
Given: Third proportion to $p$ and $q$ is $112$.
$\implies$ $p$, $q$ and $112$ are in continued proportion.
i.e. $\;$ $p : q = q : 112$
i.e. $\;$ $q^2 = 112 \times p$
$\implies$ $p = \dfrac{q^2}{112}$ $\;\;\; \cdots \; (2)$
Substituting the value of $p$ from equation $(2)$ in equation $(1)$ we have,
$196 = \dfrac{q^3}{112}$
i.e. $\;$ $q^3 = 196 \times 112 = 21952$ $\implies$ $q = \sqrt[3]{21952} = 28$
Substituting the value of $q$ in equation $(2)$ we get,
$p = \dfrac{28^2}{112} = \dfrac{784}{112} = 7$
$\therefore \;$ $p = 7$, $\;$ $q = 28$