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.
Hence, Common ratio = 3