If one geometric mean $G$ and two arithmetic means $A_{1}$ and $A_{2}$ are inserted between two distinct positive numbers, then $(\frac{2 A_{1} - A_{2}}{G}) (\frac{2 A_{2} - A_{1}}{G})$ is equal to

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Solution

The correct option is B $1$
Let two distinct positive numbers be $a$ and $b .$
Then $G^{2} = a b$
Let common difference of the A.P. $a, A_{1}, A_{2}, b$ be $d .$
We have, $2 A_{1} - A_{2} = 2 (a + d) - (a + 2 d) = a$
and $2 A_{2} - A_{1} = 2 (b - d) - (b - 2 d) = b$
Thus, $(\frac{2 A_{1} - A_{2}}{G}) (\frac{2 A_{2} - A_{1}}{G}) = \frac{a b}{a b} = 1$

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Similar questions

G

A_{1}

A_{2}

(2 A_{1} - A_{2}) (2 A_{2} - A_{1})

A_{1}

A_{2}

G^{2} = (2 A_{1} - A_{2}) (2 A_{2} - A_{1})

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\frac{A_{1} + A_{2}}{G_{1} G_{2}} =

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