If the continued product of three numbers in GP is 216 and the sum of their products in pairs is 156, then find the sum of three numbers.

Explanation

Here, product of numbers in GP is given.
`therefore` Let the three numbers be `(a)/(r ),a,ar`.
Then, `(a)/(r )*a*ar=216`
`implies a^(3)=216`
`therefore " " a=6`
Sum of the products on pairs =156
`implies (a)/(r)*a+a*ar+ar*(a)/(r)=156`
`implies a^(2)((1)/(r)+r+1)=156 implies 36((1+r^(2)+r)/r)=156`
`implies 3((1+r+r^(2))/(r))=13 implies 3r^(2)-10r+3=0`
`implies (3r-1)(r-3)=0 implies r=(1)/(3) " or "r=3`
Putting the values of a and r, the required numbers are `18,6,2`or `2,6,18`. Hence, the sum of numbers is 26.