DESMOND00766GO
Answered

If k + 1, 2k - 1 and 3k + 1 are three consecutive terms of a GP, find the value of the common ratio​

Answer :

MATHSTUDENT55GO

Answer:

r = -1 or r = 5/3

Step-by-step explanation:

(2k - 1)/(k + 1) = (3k + 1)/(2k - 1)

(2k - 1)^2 = (k + 1)(3k + 1)

4k^2 - 4k + 1 = 3k^2 + 4k + 1

k^2 - 8k = 0

k(k - 8) = 0

k = 0 or k = 8

For k = 0,

(2k - 1)/(k + 1) = -1/1 = 1

For k = 8,

(2k - 1)/(k + 1) = 15/9 = 5/3

Go Question: Other Questions