Find two numbers whose A.M. exceeds their G.M. by `1/2` and their H.M. by `25/26`.

#### Solution

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" + 1/2, "A" = "H" + 25/26`

∴ G = `"A" - 1/2, "H" = "A" - 25/26` ...(i)

Now, G^{2} = AH

`("A" - 1/2)^2 = "A"("A" - 25/26)`

∴ `"A"^2 - "A" + 1/4 = "A"^2 - 25/26"A"`

∴ `"A"- 25/26 "A" = 1/4`

∴ `1/26"A" = 1/4`

∴ A = `13/2` ...(ii)

∴ G = 6 ...[From (i) and (ii)]

∴ `"a + b"/2 = 13/2 and sqrt("ab")` = 6

∴ a + b = 13,

∴ b = 13 – a ...(iii)

and ab = 36

∴ a(13 – a) = 36 ...[From (iii)]

∴ a^{2} – 13a + 36 = 0

∴ (a – 4)(a – 9) = 0

∴ a = 4 or a = 9

When a = 4, b = 13 – 4 = 9

When a = 9, b = 13 – 9 = 4

∴ the two numbers are 4 and 9.