Latest Maths Olympiad Aptitude Question SOLUTION: three numbers in GP are such that their product is 216 if the sum of product in pairs is 156 then the numbers are

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(#M40039690) MATHS OLYMPIAD Numerical Ability question solve Keep an EYE

three numbers in GP are such that their product is 216 if the sum of product in pairs is 156 then the numbers are

Asked In Maths Olympiad Harsh Poddar (9 years ago)

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Narasapu Hari prasad  (9 years ago)

ans is 2,618
a=2,r=3

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Vaishnav  (7 years ago)

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}

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