# Find the numbers the geometric mean of two numbers is 8 and their harmonic mean is 6.4

Find the numbers the geometric mean of two numbers is 8 and their harmonic mean is 6.4
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Elliott Rollins
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, $\frac{1}{a}+\frac{a}{64}=2×\frac{5}{32}=\frac{5}{16}$
and multiplying each term by 64a we get
$64+{a}^{2}=20a$
${a}^{2}-20a+64=0$
${a}^{2}-16a-4a+64=0$
$a\left(a-16\right)-4\left(a-16\right)=0$
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,