Find the value of $x$ for which $(5 x + 2), (4 x - 1) a n d (x + 2)$ are in A.P

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Solution

given,

$5 x + 2, 4 x - 1, x + 2$ are in A.P.

$\Rightarrow 4 x - 1 - 5 x - 2 = x + 2 - 4 x + 1$

$- x - 3 = - 3 x + 3$

$3 x - x = 3 + 3$

$2 x = 6$

$∴ x = 3$

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