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

Solution :-

Let the bigger number be 'x' and the smaller number be 'y'.

Arithmetic Mean = Geometric Mean + 2

Arithmetic Mean of x and y = (x + y)/2

Geometric Mean = √xy + 2

⇒ x + y = 2√xy + 2

⇒ x + y = 2√xy + 4 .........(1)

Harmonic mean = x/5 (Given)

⇒ 2xy/x + y = x/5

⇒ 2y*5 = x + y

⇒ 10y = x + y

⇒ 10y - y = x

⇒ x = 9y ...........(2)

substituting (2) in (1)

⇒ 9y + y = 2√xy + 4

⇒ 10y = 2√9y*y + 4

⇒ 10y = 2√9y² + 4 

⇒ 10y = 2*3y + 4

⇒ 10y = 6y + 4

⇒ 10y - 6y = 4

⇒ 4y = 4

⇒ y = 4/4

⇒ y = 1

Substituting y = 1 in (2)

x = 9y

x = 9*1

x = 9

So, two numbers are 9 and 1

Answer.