# The geometric mean of two numbers is 8 and their harmonic mean is 6.4. What are the numbers?

Nov 29, 2016

Numbers are $4$ and $16$,

#### Explanation:

Let the one number be $a$ and as the geometric mean is $8$, product of two numbers is ${8}^{2} = 64$.

Hence, other number is $\frac{64}{a}$

Now as harmonic mean of $a$ and $\frac{64}{a}$ is $6.4$,

it arithmetic mean of $\frac{1}{a}$ and $\frac{a}{64}$ is $\frac{1}{6.4} = \frac{10}{64} = \frac{5}{32}$

hence, $\frac{1}{a} + \frac{a}{64} = 2 \times \frac{5}{32} = \frac{5}{16}$

and multiplying each term by $64 a$ we get

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

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

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

or $a \left(a - 16\right) - 4 \left(a - 16\right) = 0$

i.e. $\left(a - 4\right) \left(a - 16\right) = 0$

Hence $a$ is $4$ or $16$.

If $a = 4$, other number is $\frac{64}{4} = 16$ and if $a = 16$, other number is $\frac{64}{16} = 4$

Hence numbers are $4$ and $16$,