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Question
If the A.M. of two numbers exceeds their G.M. by 2 and their H.M. by `18/5`, find the numbers.
Solution
Let the required numbers be a and b and A, G, H be their A.M., G.M., H.M. respectively.
Then A = `("a" + "b")/2`, G = `sqrt("ab")` and H = `(2"ab")/("a" + "b")`
Now, A – G = 2 ...(1)
and A – H = `18/5` ...(2)
Subtracting (2) from (1), we get,
– G + H = `-8/5`
∴ H = `"G" - 8/5`
From (1), A = G + 2
∵ G2 = A.H.
∴ G2 = `("G" + 2)("G" - 8/5)`
= `"G"^2 + 2/5"G" - 16/5`
∴ 0 = `2/5"G" - 16/5`
∴ G = 8
∴ A = G + 2 gives A = 10
∴ `sqrt("ab")` = 8 and `("a" + "b")/2` = 10
∴ ab = 64, i.e., b = `64/"a"`
and a + b = 20
∴ `"a" + 64/"a"` = 20
∴ a2 + 64 = 20a
∴ a2 – 20a + 64 = 0
∴ (a – 16)(a – 4) = 0
∴ a – 16 = 0 or a – 4 = 0
∴ a = 16 or a = 4
When a = 16, a + b = 20 gives b = 4
When a = 4, a + b = 20 gives b = 16
Hence, the required numbers are 4 and 16.
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