Question

the continude product of tjree numbers of a GP is 216. Sum of product of them in pair is 156 find the numbers

Answer 1

i) Let the 3 numbers in GP be a/r, a & ar

ii) So their product is: (a/r)*(a)*(ar) = a^3 = 216; ==> a = 6

iii) As such the 3 numbers are: 6/r, 6, 6r

iv) As per the second constraint: (6/r)*(6) + 6*6r + (6r)*(6/r) = 156

Taking LCM, simplifying and rearranging, 3r² - 10r + 3 = 0
Factorizing, (3r - 1)(r - 3) = 0
==> Either r = 3 or r = 1/3

As both are feasible solutions, substituting these in (iii), we get
the 3 numbers as: {2, 6, 18} or {18, 6, 2}

EDIT:
In the final answer, the second triplet was {18, 6, 1}, which is now corrected as
{18, 6, 2}