Question

Find two numbers whose A.M. exceeds G.M. by 7 and their H.M. by `63/5`.

Answer 1

Let a, b be the two numbers.

A = `"a + b"/2, "G" = sqrt("ab"), "H" = (2"ab")/"a + b"`

According to the given conditions,

A = G + 7, A = H + `63/5`

∴ G = A – 7,                     ...(i)

H = `"A" - 63/5`

Now, G² = AH

∴ (A – 7)² = `"A"("A" - 63/5)`

∴ A² – 14A + 49 = `"A"^2 - (63"A")/5`

∴ `14"A" - (63"A")/5` = 49

∴ `(7"A")/5` = 49

∴  A = 35

∴ `"a + b"/2` = 35

∴ a + b = 70
∴ b = 70 – a               ...(ii)
G = A – 7                   ...[From (i)]
= 35 – 7
∴ G = 28
∴ `sqrt("ab")` = 28
∴ ab = 28² = 784
∴ a(70 – a) = 784    ...[From (ii)]
∴ 70a – a² = 784
∴ a² – 70a + 784
∴ a² – 56a – 14a + 784 = 0
∴ (a – 56) (a – 14) = 0
∴ a = 14 or a = 56
When a = 14, b = 70 – 14 = 14
When a = 56, b = 70 – 56 = 14
∴ the two numbers are 14 and 56.