We have been given that 2^nd, 3^rd and 6^th 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.

2^nd, 3^rd 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 2^nd and 6^th 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