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in Geometric Progressions by (15.7k points)

If 2nd, 3rd and 6th terms of an AP are the three consecutive terms of a GP then find the common ratio of the GP.

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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

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