If A and G are arithmetic and geometric means | vedclass

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Question

(b) ${x^2} - 2Ax + {G^2} = 0$&hellip;..(i)

Let $a,b$ are the two numbers whose $A.M.$ is $A$ and $G.M.$ is $G.$

$	herefore $$A = (a + b)/2,{G^2

Vedclass · Accepted Answer

$A > G$

Vedclass · Answer

(b) ${x^2} - 2Ax + {G^2} = 0$&hellip;..(i)

Let $a,b$ are the two numbers whose $A.M.$ is $A$ and $G.M.$ is $G.$

$	herefore $$A = (a + b)/2,{G^2} = ab$

From (i), ${x^2} - (a + b)x + ab = 0$

==> ${x^2} - ax - bx + ab = 0$$ \Rightarrow x(x - a) - b(x - a) = 0$

==> $(x - a)(x - b) = 0$

$	herefore a,\,b$ are the roots of the equation

$	herefore $ $A - G = \frac{{a + b}}{2} - \sqrt {ab} $

$ = \frac{1}{2}[(a + b) - 2\sqrt {ab} ]$

$ = \frac{1}{2}{(\sqrt a - \sqrt b )^2} > 0$

$\Rightarrow $$A > G$.

If $A$ and $G$ are arithmetic and geometric means and ${x^2} - 2Ax + {G^2} = 0$, then

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