If $k - 2, 2 k + 1$ and $6 k + 3$ are in G.P., then the value of $k$ is $(k > 0)$

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Solution

The correct option is A $7$
Since, $k - 2, 2 k + 1 a n d 6 k + 3$ are in GP.
$\Rightarrow (2 k + 1)^{2} = (k - 2) (6 k + 3)$
$\Rightarrow 4 k^{2} + 4 k + 1 = 6 k^{2} + 3 k - 12 k - 6$
$\Rightarrow 2 k^{2} - 13 k - 7 = 0$
$\Rightarrow 2 k^{2} - 14 k + k - 7 = 0$
$\Rightarrow (2 k + 1) (k - 7) = 0$
$\Rightarrow k = - \frac{1}{2}, 7$
There should be two value $- \frac{1}{2}, 7$

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If k−2,2k+1 and 6k+3 are in G.P., then the value of k is (k>0)

The correct option is A 7Since, k−2,2k+1 and 6k+3 are in GP.⇒(2k+1)2=(k−2)(6k+3)⇒4k2+4k+1=6k2+3k−12k−6⇒2k2−13k−7=0⇒2k2−14k+k−7=0⇒(2k+1)(k−7)=0⇒k=−12,7There should be two value −12,7

If $k - 2, 2 k + 1$ and $6 k + 3$ are in G.P., then the value of $k$ is $(k > 0)$