Question

Insert two numbers between `1/4` and `1/3` so that the resulting sequence is a H.P.

Answer 1

Let the required numbers be H₁ and H₂.

Then `1/4, "H"_1, "H"_2, 1/3` are in H.P.

∴ `4, 1/"H"_1, 1/"H"_2, 3` are in A.P.

∴ t₁ = 4, t₂ = `1/"H"_1`, t₃ = `1/"H"_2`, t₄ = 3

If a is the first term and d is the common difference of the AP., then a = t₁ = 4 and t₄ = a + (4 – 1)d = 3

∴ 4 + 3d = 3

∴ 3d = – 1

∴ d = `-1/3`

∴ `1/"H"_1 = "t"_2 = "a" + "d" = 4 - 1/3 = 11/3`

∴ H₁ = `3/11`

and `1/"H"_2 = "t"_3 = "a"+ 2"d" = 4 + 2(-1/3) = 10/3`

∴ H₂ = `3/10`

Hence, the required numbers are `3/11 and 3/10`.