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If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a ge...

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

A.

0, 8

B.

-1, 53

C.

2, 3

D.

1, -1

Correct answer is B

2k1k+1=3k+12k1

(k+1)(3k+1)=(2k1)(2k1)

3k2+4k+1=4k24k+1

4k23k24k4k+11=0

k28k=0

k(k8)=0

The terms of the sequence given k = 0: (1, -1, 1)

\implies \text{The common ratio r = -1}

The terms of the sequence given k = 8: (9, 15, 25)

\implies \text{The common ratio r = } \frac{5}{3}

The possible values of the common ratio are -1 and \frac{5}{3}.