Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is

Solution :

Let a, ar, \(ar^2\) are in GP (r > 1)

According to the question, a, 2ar, \(ar^2\) in AP.

\(\implies\)  4ar = a + \(ar^2\)

\(\implies\) \(r^2\) – 4r + 1 = 0

\(\implies\) r = \(2 \pm \sqrt{3}\)

Hence, r = \(2 + \sqrt{3}\)    [ \(\because\)  AP is increasing]

Similar Questions

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……

Leave a Comment

Your email address will not be published. Required fields are marked *