Question

Find two numbers whose Arithmetic mean exceeds the geometric mean by 2 and whose harmonic mean is one fifth of the larger number

Answer 1

1×2/5
will give your answer

Answer 2

Answer:

9 and 1

Step-by-step explanation:

Let the two numbers be a and b such that a>b

Arithmetic mean formula is A.M $=\frac{a+b}{2}$

Geometric mean formula is G.M =$\sqrt{ab}$

A.M exceeds their G.M by 2

A.M = G.M +2

$\frac{a+ b}{2}=\sqrt{ab} + 2$

$a+b = 2(\sqrt{ab}+2)$------- 1

H.M is one – fifth larger number .

H.M =$\frac{1}{5} a$

$\frac{2ab}{a+b} = \frac{1}{5} a$

$\frac{2b}{a + b}=\frac{1}{5}$

$a + b = 10b$ ----------------->2

$a + b- 10b = 0$

$a-9b = 0$

$a = 9b$

Substitute a = 9b in Equation 1

$9b+b = 2(\sqrt{9b \times b}+2)$

$10b = 2 (\sqrt{9b^2}$

$10b = 2( 3b + 2 )$

$10b = 6b + 4$

$10b-6b = 4$

$4b = 4$

$b = 1$

Substitute the value of b in 2 to get value of a

$a + b = 10b$

Put b = 1

$a + (1) = 10 (1)$

$a + 1 = 10$

$a = 10-1$

$a = 9$

Hence the two numbers are 9 and 1