Determine the value of $k$ such that $k + 2, 4 k - 6, 3 k - 2$ are in A.P.

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Solution

Given: $k + 2, 4 k - 6, 3 k - 2$ are in A.P.

We know that , if the terms are in A.P the difference between terms is same.

$\Rightarrow∴ t_{2} - t_{1} = t_{3} - t_{2}$

$\Rightarrow 4 k - 6 - (k + 2) = 3 k - 2 - (4 k - 6)$

$\Rightarrow 4 k - 6 - k - 2 = 3 k - 2 - 4 k + 6$

$\Rightarrow 3 k - 8 = - k + 4$

$\Rightarrow 3 k + k = 4 + 8$

$\Rightarrow 4 k = 12$

$\Rightarrow k = \frac{12}{4} = 3$

$∴ k = 3$

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