If one G.M. 'G' and two arithmetic means p and q be inserted between any two number then $G^{2}$ =

(2 p - q) (2 q - p)

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(2 p - q) (q - 2 p)

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\frac{2 p + q}{2 q + p}

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\frac{2 p + q}{q + 2 p}

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Solution

The correct option is A $(2 p - q) (2 q - p)$
$G = \sqrt{a b}$
$a, p, q, b$ are in $A P$
$b = a + 3 d \Rightarrow d = \frac{b - a}{3}$
$p = a + \frac{b - a}{3} = \frac{2 a + b}{3}$
$q = p + d = \frac{2 a + b}{3} + \frac{b - a}{3}$
$= \frac{2 b + a}{3}$
$2 p - q = \frac{3 a}{3} = a$
$p - 2 q = - b$
$(2 p - q) (p - 2 q) = - a b$
$(2 p - q) [- (2 q - p)] = - G^{2}$
$(2 p - q) (2 q - p) = G^{2}$
$\Rightarrow G^{2} = (2 p - q) (2 q - p)$

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If one geometric mean G and two arithemtic means p and q be inserted between two numbers, then $G^{2}$ is equal to:

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