The AM and GM of two numbers are 17 and 8 res | vedclass

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(C) Given $AM = 17$ and $GM = 8$.Let the two numbers be $a$ and $b$.We know that $AM = \frac{a+b}{2} = 17 \implies a+b = 34$ $(1)$.We know that $GM =

Vedclass · Accepted Answer

$\frac{64}{17}$

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(C) Given $AM = 17$ and $GM = 8$.Let the two numbers be $a$ and $b$.We know that $AM = \frac{a+b}{2} = 17 \implies a+b = 34$ $(1)$.We know that $GM = \sqrt{ab} = 8 \implies ab = 64$ $(2)$.Using the relation $(a-b)^2 = (a+b)^2 - 4ab$,we get:$(a-b)^2 = (34)^2 - 4(64) = 1156 - 256 = 900$.Thus,$a-b = 30$ $(3)$.Solving $(1)$ and $(3)$,we get $2a = 64 \implies a = 32$ and $b = 2$.The Harmonic Mean $(HM)$ is given by $HM = \frac{2ab}{a+b} = \frac{2(64)}{34} = \frac{128}{34} = \frac{64}{17}$.The $AP$ is $2, 17, 32$.The $GP$ is $2, 8, 32$.The $HP$ is $2, \frac{64}{17}, 32$.

The $AM$ and $GM$ of two numbers are $17$ and $8$ respectively. Find their Harmonic Mean $(HM)$ and write the $AP, GP,$ and $HP$.

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