Question

Q # 54 - If H.M and A.M between two numbers are 4 and 9/2 respectively, then the numbers are​

Answer 1

Answer:-

Let the two numbers be a and b.

Given:

HM of a , b = 4

Harmonic mean between a , b = 2ab/a + b

So,

⟹ 2ab/a + b = 4

⟹ 2ab = 4(a + b)

⟹ 2ab/4 = a + b

⟹ ab/2 = a + b -- equation (1).

Also given that;

AM of a , b = 9/2.

We know that;

Arithmetic mean = (a + b)/2

So,

⟹ (a + b)/2 = 9/2

⟹ a + b = (9/2)(2)

⟹ a + b = 9 -- equation (2).

Substitute the value of a + b in equation (1).

⟹ ab/2 = 9

⟹ ab = 18

⟹ a = 18/b

Now, substitute a = 18/b in equation (2).

⟹ (18/b) + b = 9

⟹ (18 + b²)/b = 9

⟹ b² + 18 = 9b

⟹ b² - 9b + 18 = 0

⟹ b² - 6b - 3b + 18 = 0

⟹ b(b - 6) - 3(b - 6) = 0

⟹ (b - 3)(b - 6) = 0

⟹ b = (3 , 6)

Substitute b = 3 in equation (2).

⟹ a + 3 = 9

⟹ a = 9 - 3

⟹ a = 6

(or)

Substitute b = 6 in equation (2).

⟹ a + 6 = 9

⟹ a = 9 - 3

⟹ a = 6

∴ The two numbers are 3 , 6 (a ≠ b).