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The fourth, seventh and tenth terms of a GP are $p, q, r$ respectively, then

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$p^{2} = q^{2} + r^{2}$

$q^{2} = p r$

$p^{2} = q r$

$p q r + p q + 1 = 0$

answer is B.

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Detailed Solution

$G_{4} = p, G_{7} = q, G_{10} = r$

$\begin{matrix} t_{4} = p, & t_{7} = q, & t_{10} = r \\ a . r^{3} = p, & a . r^{6} = q, & a . r^{9} = r \end{matrix}$

$(a . r^{3}) (a . r^{9}) = a^{2} r^{12}$

$= {(a . r^{6})}^{2}$

$= q^{2}$

$q^{2} = p r$

$p, q, r$ are in GP then $q^{2} = p r$

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