Correct Answer - Option 4 : 3/2

__Concept:__

If a, ar, ar2, ar3,.....,arn-1 are in GP then nth term of GP is given by Tn = arn-1

If a_{1}, a_{2}, a_{3}, a_{4} are in GP, then common ratio of GP is given by, \( r=\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}\)

__Calculation:__

Lets three terms of GP be a/r, a, ar

Given: The product of first three terms of GP = 216 and their sum is 19

\(⇒ \rm \frac{a}{r}×{a}×{ar}=125\) ⇒ a^{3} = 216 ⇒ a = 6

According to second statement,

\(⇒ \rm \frac{a}{r}+{a}+{ar}=19⇒ a\left (\frac{1}{r}+1+r \right )=19\)

Put the value of a = 6 in the above equation

\(⇒ \rm 6\left (\frac{1}{r}+1+r \right )=19\)

⇒ 6 + 6r + 6r^{2} = 19r ⇒ 6r^{2} - 13r + 6 = 0

⇒ (3r - 2) × (2r - 3) = 0

⇒ r = 2/3 or 3/2

As it is given that r > 1

Hence common ratio r = 3/2.