If $p, q, r$ are three distinct real numbers in A.P., then $p^{3} + r^{3}$ equals

p q r

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- 3 p q r

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- 6 p q r

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None of these

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Solution

The correct option is D None of these
Let us take $p, q$ and $r$ to be $a - d, a$ and $a + d$ as $3$ distinct numbers in AP.
Then $({a - d)}^{3} + ({a + d)}^{3}$
$= 2 a (a^{2} + d^{2} - 2 a d + a^{2} + d^{2} + 2 a d - a^{2} + d^{2})$
$= 2 a (a^{2} + {3 d}^{2})$
Writing in terms of $p, q$ and $r$ , we get
$p^{3} + r^{3} = (p + r) (p^{2} - p r + r^{2})$
$= (2 q) (p^{2} - p r + r^{2})$
which does not match with the given options.
Hence, option D is correct.

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