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# If A be the A.M. and G be the G.M. between two numbers. Show that the numbers are given by $A \pm \sqrt {{A^2} - {G^2}}$

Last updated date: 17th Apr 2024
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Hint: We know that A.M. is arithmetic mean, and G.M. is geometric mean, and A.M and G.M. for two numbers say ‘a’ and ‘b’ will be $\dfrac{{a + b}}{2}$ and $\sqrt[{}]{{ab}}$ respectively. We will make a quadratic equation using it, and the roots of the equation gives the value of the numbers which we have to prove.

Complete step-by-step solution:
Given, A and G are A.M and G.M between two numbers. Let the two numbers be ‘a’ and ‘b’, we know that A.M between two numbers is the average of two numbers and G.M between two numbers is the square root of the product of the numbers.
Then, $A = \dfrac{{a + b}}{2}$ and $G = \sqrt {ab}$
Simplifying them, we get:
$\Rightarrow a + b = 2A\,\,\,\,\,---------- equation\,1$
and,
$\Rightarrow ab = {G^2}\,\,\,\,\, -------- equation\,2$
$\Rightarrow {\left( {a - b} \right)^2} = {\left( {a + b} \right)^2} - 4ab \\ \Rightarrow {\left( {a - b} \right)^2} = {\left( {2A} \right)^2} - 4{G^2} \\ \Rightarrow {\left( {a - b} \right)^2} = 4\left( {{A^2} - {G^2}} \right) \\ \Rightarrow \left( {a - b} \right) = \pm 2\sqrt {{A^2} - {G^2}} \,\,--- equation\,3$
Taking equation 1 and equation 2 we get,
$\Rightarrow a - b = 2\sqrt {{A^2} - {G^2}} \\ \Rightarrow a + b = 2A$
$\Rightarrow a = A + \sqrt {{A^2} - {G^2}}$
$\Rightarrow a + b = 2A \\ \Rightarrow A + \sqrt {{A^2} - {G^2}} + b = 2A$
$\Rightarrow b = A - \sqrt {{A^2} - {G^2}}$
$= A - \sqrt {\left( {A + G} \right)\left( {A - G} \right)}$
Hence, the numbers are $A \pm \sqrt {{A^2} - {G^2}}$