Question

Arithmetic mean of two numbers is 75 and their Geometric mean is 21 then what are the numbers​

Answer 1

Given:

The arithmetic mean of two numbers is 75

The geometric mean of two numbers is 21

To find:

The numbers

Solution:

Let "a" & "b" be the two positive numbers.

We know that,

If a and b are two positive numbers then

$\boxed{\bold{A.M. = \frac{a+b}{2} }}$

$\boxed{\bold{G.M. = \sqrt{ab} }}$

∵ A.M. = 75

∴ $\frac{a+b}{2} = 75$

$\implies a+b = 150$ ..... (i)

∵ G.M. = 21

∴ $\sqrt{ab} = 21$

squaring on both sides

$\implies ab = 21^2$

$\implies ab = 441$

$\implies a = \frac{441}{b}$ ..... (ii)

Now, substituting (ii) in (w), e get

$\frac{441}{b} + b = 150$

$\implies \frac{441 \:+\: b^2}{b} = 150$

$\implies 441 \:+\: b^2 = 150b$

$\implies b^2 -150b +441=0$

$\implies b^2 -147b - 3b +441=0$

$\implies b(b - 147) - 3(b - 147) = 0$

$\implies (b - 147)(b - 3) = 0$

$\implies b = 147 \:or\: 3$

So,

When b = 147

Then substituting in (ii), we get ⇒ a = $\frac{441}{147}$ = 3

and

When b = 3

Then substituting in (ii), we get ⇒ a = $\frac{441}{3}$ = 147

Thus, the two numbers are 147 and 3.

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Answer 2

Answer:

Step-by-step explanation: