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, (frac{5}{3}) 
C.
2, 3 
D.
1, 1
Correct Answer: Option B
Explanation
(frac{2k – 1}{k + 1} = frac{3k + 1}{2k – 1})
((k + 1)(3k + 1) = (2k – 1)(2k – 1))
(3k^{2} + 4k + 1 = 4k^{2} – 4k + 1)
(4k^{2} – 3k^{2} – 4k – 4k + 1 – 1 = 0)
(k^{2} – 8k = 0)
(k(k – 8) = 0)
(therefore text{k = 0 or 8})
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}).