Question

if the arithmetic mean of two number is 4.5 and their harmonic mean is 4 then find numbers

Answer 1

let numbers be a and b, then

Arithmetic Mean :

a+b / 2 = 4.5 or a+b = 9----I

Harmonic Mean :  2ab / (a+b) = 4

or ab=18 ----II

We have the sum and product from I and II, we need to find the a and b.

We will do it using Algebraic identity, (a+b)^{2}  - (a-b)^{2}  = 4ab

9^{2} - (a-b)^{2} = 4(18)

a-b=3 ----III

Solve I and III for value of a and b

we get a=6 and b=3

Answer 2

Answer:

The two numbers will be 3 and 6.

Step-by-step explanation:

We have given the arithmetic mean = 4.5

The harmonic mean of two numbers = 4

Consider that 'a' and 'b' are the two numbers.

The arithmetic mean of a and b $=\frac{a+b}{2}$

$4.5=\frac{a+b}{2}$

$a+b=9$

$b=9-a$                                                                       .................(1)

Harmonic mean of two numbers is the reciprocal of the arithmetic mean of reciprocals of those two numbers.

The harmonic mean of a and b $=\frac{2ab}{a+b}$

$4=\frac{2ab}{a+b}$

$ab=2(a+b)$

$ab=2(9)\\ab=18$

Substitute the value of 'b' from equation (1);

$a(9-a)=18$

$9a-a^2-18=0\\a^2-9a+18=0$

$a^2-3a-6a+18=0$

$a(a-3)-6(a-3)=0$

$(a-3)(a-6)=0$

$a=3,6$

Substitute the value of a in equation (1), we get;

For $a=3, \; b=9-3 =6$

For a=6, b = 9-6 = 3

Therefore, the two numbers will be 3 and 6.