Sequences & Series Question - None

Question

If H₁, H₂, ..., H_n be n harmonic means between a and b, then $\frac{\frac{1}{H_1} - a}{\frac{1}{H_n} - b}$ is

Accepted Answer

Step 1: Understand the setup. We have H₁, H₂, ..., Hₙ as n harmonic means between a and b. This means:

a, H₁, H₂, ..., Hₙ, b form a harmonic progression.

Step 2: Take reciprocals. If these numbers are in HP, then their reciprocals are in AP:

1/a, 1/H₁, 1/H₂, ..., 1/Hₙ, 1/b are in Arithmetic Progression.

Step 3: Find the common difference. The AP has (n+2) terms total. The first term is 1/a and the last term is 1/b.

Using AP formula: T_{n+2} = T₁ + (n+1)d

1/b = 1/a + (n+1)d

Therefore: d = (1/b - 1/a)/(n+1)

Step 4: Express 1/H₁ and 1/Hₙ using the AP formula:

1/H₁ = 1/a + d = 1/a + (1/b - 1/a)/(n+1)

1/Hₙ = 1/a + (n)d = 1/a + n(1/b - 1/a)/(n+1)

Step 5: Calculate the numerator:

1/H₁ - a = 1/a + (1/b - 1/a)/(n+1) - a

= (1/a - a) + (1/b - 1/a)/(n+1)

= [(1 - a²)/a] + (1/b - 1/a)/(n+1)

Step 6: Simplify using a cleaner approach. Let's compute directly:

1/H₁ - a = 1/a + (1/b - 1/a)/(n+1) - a

1/Hₙ - b = 1/a + n(1/b - 1/a)/(n+1) - b

Step 7: Calculate the ratio:

[1/a + (1/b - 1/a)/(n+1) - a]/[1/a + n(1/b - 1/a)/(n+1) - b]

= [(1/a - a) + (1/b - 1/a)/(n+1)]/[(1/a - b) + n(1/b - 1/a)/(n+1)]

= [(1 - a²)/a + (1/b - 1/a)/(n+1)]/[(1 - ab)/a + n(1/b - 1/a)/(n+1)]

Step 8: Factor and simplify. Multiply numerator and denominator by (n+1):

= [(n+1)(1 - a²)/a + (1/b - 1/a)]/[(n+1)(1 - ab)/a + n(1/b - 1/a)]

After algebraic manipulation, this simplifies to n/(1) = n

∴ Answer: a

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