KnowledgeBoat Logo
|

Mathematics

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

AP GP

106 Likes

Answer

In a G.P.,

 any term  preceding term = common ratio .∴2(k+1)k=3(k+1)2(k+1) (Eq 1) ⇒(2k+2)(2k+2)=k(3k+3)⇒4k2+4k+4k+4=3k2+3k⇒4k2−3k2+8k−3k+4=0⇒k2+5k+4=0⇒k2+k+4k+4=0⇒k(k+1)+4(k+1)=0⇒(k+1)(k+4)=\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+1k + 1 ≠ 0 as that will not satisfy Eq 1

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

Hence, Option 2 is the correct option.

Answered By

8 Likes

Related Questions