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

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Answer 1 · 2016-11-29T09:14:15

Numbers are $4$ $4$ and $16$ $16$ ,

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

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

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

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

hence, $\frac{1}{a} + \frac{a}{64} = 2 \times \frac{5}{32} = \frac{5}{16}$ $1/a+a/64=2xx5/32=5/16$

and multiplying each term by $64 a$ $64a$ we get

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

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

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

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

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

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

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

Hence numbers are $4$ $4$ and $16$ $16$ ,