Question

Find two numbers such that the mean proportional between them is 12 and the third proportional to them is 96. [3 Marks]

Answer 1

Step: Find the numbers

Let a and b be the two numbers, whose mean proportional is 12

$\therefore ab={12}^2$ [1 Mark]

$\Rightarrow ab=144$

$\Rightarrow b=\frac{144}{a}$ ...(i)

a,b and 96 in proportion

$\therefore \frac{a}{b}=\frac{b}{96}$

$\Rightarrow b^2=96a$ [1 Mark]

$\Rightarrow {\left(\frac{144}{a}\right)}^2=96a$

​​​​​​​$\Rightarrow \frac{144\times 144}{a^2}=96a$

​​​​​​​$\Rightarrow 144\times 144=96\times a^3$ [1 Mark]

​​​​​​​$\Rightarrow a^3=\frac{144\times 144}{96}=\frac{20736}{96}$

​​​​​​​$\Rightarrow a^3=216=6\times 6\times 6$

​​​​​​​$\therefore a=\sqrt[3]{36^3}=6$

From (i), $b=\frac{144}{6}=24$

Hence, the values of a and b are 6 and 24 respectively. [1 Mark]