If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is

1. -1

2. -4

3. 1

4. 4

In a G.P.,

$\dfrac{\text{ any term }}{\text{ preceding term }} = \text{ common ratio }. \\[1em] \therefore \dfrac{2(k + 1)}{k} = \dfrac{3(k + 1)}{2(k + 1)} \text{ (Eq 1) } \\[1em] \Rightarrow (2k + 2)(2k + 2) = k(3k + 3) \\[1em] \Rightarrow 4k^2 + 4k + 4k + 4 = 3k^2 + 3k \\[1em] \Rightarrow 4k^2 - 3k^2 + 8k - 3k + 4 = 0 \\[1em] \Rightarrow k^2 + 5k + 4 = 0 \\[1em] \Rightarrow k^2 + k + 4k + 4 = 0 \\[1em] \Rightarrow k(k + 1) + 4(k + 1) = 0 \\[1em] \Rightarrow (k + 1)(k + 4) =$

But $k + 1$ ≠ 0 as that will not satisfy Eq 1

$\Rightarrow k + 4 = 0 \\[1em] \Rightarrow k = -4.$

Hence, Option 2 is the correct option.