If $p - 1, p + 3, 3 p - 1$ are in A.P. then $p$ is equal to:

4

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2

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- 2

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Solution

The correct option is A $4$
Given:
$p - 1, p + 3, 3 p - 1$ are in A.P.
$∴ 2 (p + 3) = p - 1 + 3 p - 1$
$\Rightarrow 2 p + 6 = 4 p - 2$
$\Rightarrow 2 p = 8$
$∴ p = 4$

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