Find the value of $x$ for which $(5 x + 2), (4 x - 1)$ and $(x + 2)$ are in AP.

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Solution

Given, $(5 x + 2), (4 x - 1)$ and $(x + 2)$ are in AP.
Then,
$2 (4 x - 1) = {(5 x + 2) + (x + 2)}$
or, $8 x - 2 = 6 x + 4$
or, $2 x = 6$
or, $x = 3$ .

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