Question

Three numbers are in continued proportion, whose mean proportional is $12$ and the sum of the remaining two numbers is $26$, then find these numbers.

Answer 1

let the there numbers be

a,b,c

As they are in continued

Proportion

$b^2=ac$

mean is 12

$\therefore b=12$

So, ${12}^2=ac$

$144=ac$

$c=144/a$

Sum of two number is

$a+c=26$

$a+\frac{144}{a}=26$

Solving,

$=a^2+144-26a=0$

$=a^2-18a-8a+144=0$

$=a(a-18)-8(a-18)=0$

$=(a-18)(a-8)=0$

$\therefore a=18$ or $a=8$

$c=8$ $c=18$

$\therefore$ The numbers are :-

$8,12,18$ or $18,12,8$