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The value of ‘k’ so that $k + 2, 4 k - 6, 3 k - 2$ are in A.P is

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answer is C.

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Detailed Solution

$d = 4 k - 6 - k - 2 = 3 k - 8$

(or)

$d = 3 k - 2 - 4 k + 6 = - k + 4$

$∴ 3 k - 8 = - k + 4$

$4 k = 12$

$k = 3$

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