Answer : We have been given that 2nd, 3rd and 6th terms of an AP are the three consecutive terms of a GP

Let the three consecutive terms of the G.P. be a,ar,ar2.

Where a is the first consecutive term and r is the common ratio.

2nd, 3rd terms of the A.P. are a and ar respectively as per the question.

∴ The common difference of the A.P. = ar – a And the sixth term of the A.P. = ar2

Since the second term is a and the sixth term is ar2(In A.P.) We use the formula:t = a + (n – 1)d

∴ ar2 = a + 4(ar – a)…(the difference between 2nd and 6th term is 4(ar – a))

⇒ ar2 = a + 4ar – 4a

⇒ ar2 + 3a – 4ar = 0

⇒ a(r2 – 4r + 3) = 0

⇒ a(r – 1)(r – 3) = 0

Here, we have 3 possible options:

1)a = 0 which is not expected because all the terms of A.P. and G.P. will be 0.

2)r = 1,which is also not expected because all th terms would be equal to first term. 3)r = 3,which is the required answer.

Ans: Common ratio = 3