Question

Find two numbers whose arithmetic mean is 12.5 and geometric mean is 10

Answer 1

Let the numbers be a, b.

So, arithmetic mean =

$\frac{a + b}{2}$

⟹

$\frac{a + b}{2} = 10$

⟹

$a+b = 20$

Geometric mean =

$√ab$

⟹

$√ab = 8$

⟹

$ab = 64$

Since

${a} - b = 20 </p><p>⟹ b = 20-a</p><p>⟹ a × (20-a) = 64$

⟹

${a}^{2} -20a+64 = 0$

⟹

${a}^{2} -16a-4a+64 = 0$

⟹

$(a-16) × (a-4) =0$

⟹

$a= 16 or a= 4.$

So, the numbers are 16, 4.

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Answer 2

Answer:

Two numbers whose arithmetic mean is 12.5 and geometric mean is 10 are 20 and 5

Step-by-step explanation:

Given,

The arithmetic mean of two numbers = 12.5

The geometric mean of two numbers = 10

To find,

The numbers

Recall the concept

Then arithmetic mean of the two numbers a and b= $\frac{a+b}{2}$

The geometric mean of two numbers a and b =$\sqrt{ab}$

(a-b)² = (a+b)² - 4ab

Solution:

Let 'a' and 'b' be the two numbers

Since the arithmetic mean of two numbers = 12.5 we have

$\frac{a+b}{2}$ = 12.5

a+b = 25 -----------------(1)

Since Geometric mean of two numbers = 10

$\sqrt{ab}$  = 10

ab = 100-----------------(2)

We have, (a-b)² = (a+b)² - 4ab

(a-b)² = 25² - 4×100

=625 - 400

=225

a -b = 15 ---------------(3)

Adding (1) and (2) we get

2a = 40

a = 20

Substituting the value of 'a' in equation (1) we get

20 +b = 25

b = 5

∴ Two numbers whose arithmetic mean is 12.5 and geometric mean is 10 are a = 20 and 5

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