If 2nd, 3rd and 6th terms of an A.P. are the three consecutive terms of a G.P., then the common ratio of the G.P. is :

1. 2

2. 3

3. $\dfrac{1}{2}$

4. $\dfrac{1}{3}$

Let first term of A.P. be a and common difference be d.

2nd Term : a + d

3rd Term : a + 2d

6th Term : a + 5d

These three terms form three consecutive terms of a G.P.

Thus, a + d, a + 2d, a + 5d, …….. is the G.P.

In G.P., ratio between consecutive terms are equal.

$\Rightarrow \dfrac{a + 5d}{a + 2d} = \dfrac{a + 2d}{a + d}$

⇒ (a + 2d)² = (a + d)(a + 5d)

⇒ a² + 4ad + 4d² = a² + 5ad + ad + 5d²

⇒ a² + 4ad + 4d² = a² + 6ad + 5d²

⇒ 0 = a² - a² + 6ad - 4ad + 5d² - 4d²

⇒ 0 = 2ad + d²

⇒ d(2a + d) = 0

⇒ d = 0 or 2a + d = 0

d cannot be equal to zero as then common ratio will be equal to 1, also not in options.

⇒ 2a + d = 0

⇒ d = -2a

Substitute d = −2a :

a + d = a - 2a = -a

a + 2d = a + 2(-2a) = -3a

a + 5d = a + 5(-2a) = -9a

r = $\dfrac{-3a}{-a}$ = 3.

Hence, option 2 is the correct option.