If one geometric mean G and two arithmetic means $A_{1}$ and $A_{2}$ be inserted between two given quantities, prove that $G^{2} = (2 A_{1} - A_{2}) (2 A_{2} - A_{1})$ .

Open in App

Solution

Let a and b be two given quantities. It is given that G is the geometric mean of a and b
Therefore, $G = \sqrt{a b}$
$G^{2} = a b$
It is also given that $A_{1}, A_{2}$ are two arithmetic means between a and b. Therefore, $a, A_{1}, A_{2}, b$ is an A.P. with common difference $d = \frac{b - a}{3}$

Therefore, $A_{1} = a + d = a + \frac{b - a}{3} = \frac{2 a + b}{3}$

$A_{2} = a + 2 d = a + \frac{2 (b - a)}{3} = \frac{a + 2 b}{3}$

So, $2 A_{1} - A_{2} = 2 (\frac{2 a + b}{3}) - (\frac{a + 2 b}{3}) = a$

and $2 A_{2} - A_{1} = 2 (\frac{a + 2 b}{3}) - (\frac{2 a + b}{3}) = b$

Therefore,
$(2 A_{1} - A_{2}) (2 A_{2} - A_{1}) = a b$
$(2 A_{1} - A_{2}) (2 A_{2} - A_{1}) = G^{2}$

Suggest Corrections

Similar questions

G

A_{1}

A_{2}

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

G

A_{1}

A_{2}

(\frac{2 A_{1} - A_{2}}{G}) (\frac{2 A_{2} - A_{1}}{G})

A_{1}, A_{2}

G_{1}, G_{2} : H_{1} H_{2} = A_{1} + A_{2} : H_{1} + H_{2}

A_{1}, A_{2}

G_{1}, G_{2}

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

G

p

q

G^{2}

Join BYJU'S Learning Program