hikstac0

2022-09-09

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

Elliott Rollins

Beginner2022-09-10Added 8 answers

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 64a we get

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

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

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

$a(a-16)-4(a-16)=0$

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

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,

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 64a we get

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

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

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

$a(a-16)-4(a-16)=0$

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

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,

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