Question

If the A.M. of two numbers exceeds their G.M. by 10 and their H.M. by 16, find the numbers.

Answer 1

Let x and y are two numbers

Arithmetic mean , A.M of x and y = (x + y)/2

geometric mean , G.M of x and y = √(xy)

harmonic mean, H.M of x and y = 2xy/(x + y)


a/c to question,
AM = GM + 10 => GM = AM - 10
AM = HM + 16 => HM = AM - 16

we know, GM² = AM × HM
(AM - 10)² = AM × (AM - 16)

AM² - 20AM + 100 = AM² - 16AM

AM = 25

GM = 25 - 10 = 15

and HM = 25 - 16 = 9

so, (x + y)/2 = 25 => x + y = 50 [ as AM = 25 ]
√xy = 15 => xy = 225 [ as GM = 15]

now, (x + y)² = (x - y)² + 4xy = 50²

(x - y)² + 4 × 225 = 2500

(x - y)² = 2500 - 900 = 1600

(x - y) = ± 40

so, if x = 45 and y = 5 or x = 5 and y = 45

hence, two numbers are 5 and 45.

Answer 2

Answer:

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