Question

find G.M of two positive numbers whose A.M and H.M are 12 and 3 respectively

Answer 1

$\huge{\pink{\underline{\mathfrak{your\:answer}}}}$

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$\bold{G.M\:=\:6}$

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step-by-step explanation:

Let the two positive numbers be 'x' and 'y'.

Now,

it is given that,

Arithmetic mean, A.M = 12

$= > \frac{x + y}{2} = 12$

=> x + y = 12 × 2

=> x + y = 24..................(i)

Now,

given that,

Harmonic mean, H.M = 3

$= > \frac{2}{ \frac{1}{x} + \frac{1}{y} } = 3$

doing cross multiplication,

we get,

$= > \frac{1}{x} + \frac{1}{y} = \frac{2}{3}$

Taking L.C.M of denominators and simplifying,

we get,

$= > \frac{x + y}{xy} = \frac{2}{3}$

But, from eqn (i),

(x+y) = 24,

so putting the value,

we get,

$= > \frac{24}{xy} = \frac{2}{3}$

On cross-multiplication,

we get,

$= > xy = \frac{24 \times 3}{2}$

$= > xy = 36$

But,

we know that,

Geometric mean, G.M of two numbers is the square root of their product.

so,

G.M of 'x' and 'y' is $\sqrt{xy}$

But,

xy = 36

therefore,

G.M = $\sqrt{36}$

=> G.M = 6

Hence,

Geometric mean,G.M of the numbers = 6

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Answer 2

let x and y be positive numbers

GM= √xy

AM =(x+y)/2

HM =2xy/(x+y)

AM(HM)=xy =GM^2

now

GM^2 =(AM)(HM)

=12(3)=36=6^2

GM =6