Question

The sum of 3 numbers of a GP is 26 and their product is 216 . Find the numbers

Answer 1

Let the three numbers in Geometric Progression be : a,a*r,a*r^2

Given,(a)*(a*r)*(a*r^2)=216

=>a^3*r^3=216

=>(a*r)^3=216  

=>a*r=6

=>a=6/r

It is also given that,  a+a*r+a*r^2 = 26

(a)+(a*r)+(a*r*r)=26

Substituiting the values of 'a*r' and 'a'

6/r+6+6r=20

=>6+6r^2=20r

=>6r^2-20r+6=0

=>6r^2-2r-18r+6=0

=>2r(3r-1)-6(3r-1)=0

=>(3r-1)(2r-6)=0

So,r=1/3 , 1/3.

First number(a)=6/r=6/(1/3)=18

Second number (a*r)=18*1/3=6

Third number(a*r^2)=18*1/9=2.

Thanks!!!