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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.

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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.

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