The value of $p$ if $p, 64, p^{2}$ are in GP

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Solution

The correct option is C 16
Given series is $p, 64, p^{2}$
Condition to be in GP is
$b^{2} = a c 64^{2} = p (p^{2}) p^{3} = 4^{6} p = 4^{2} p = 16$

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