Sequences & Series Question - None

Question

If a, a₁, a₂, a₃, ..., a_2n, b are in AP and a, b₁, b₂, b₃, ..., b_2n, b are in GP and h is the HM of a and b, then

$\frac{a_1 + a_{2n}}{b_1 b_{2n}} + \frac{a_2 + a_{2n-1}}{b_2 b_{2n-1}} + \ldots + \frac{a_n + a_{n+1}}{b_n b_{n+1}}$ is equal to

Accepted Answer

Step 1: Set up the AP progression.

We have a, a₁, a₂, ..., a₂ₙ, b in AP with common difference d.

There are 2n+2 terms total. Thus: b = a + (2n+1)d, so d = (b-a)/(2n+1)

Step 2: Find the general term of the AP.

aᵢ = a + id for i = 1, 2, ..., 2n

Key property: aᵢ + a₍₂ₙ₊₁₋ᵢ₎ = (a + id) + (a + (2n+1-i)d) = 2a + (2n+1)d = a + b

Step 3: Set up the GP progression.

We have a, b₁, b₂, ..., b₂ₙ, b in GP with common ratio r.

Thus: b = ar^(2n+1), so r = (b/a)^(1/(2n+1))

Step 4: Find the general term of the GP.

bᵢ = ar^i for i = 1, 2, ..., 2n

Key property: bᵢ · b₍₂ₙ₊₁₋ᵢ₎ = (ar^i)(ar^(2n+1-i)) = a²r^(2n+1) = ab

Step 5: Evaluate each term in the sum.

Each term in the sum has the form: (aᵢ + a₍₂ₙ₊₁₋ᵢ₎)/(bᵢ · b₍₂ₙ₊₁₋ᵢ₎)

Substituting our results: (a + b)/(ab) for each term

Step 6: Calculate the total sum.

The sum contains n terms (from i=1 to i=n), so:

Sum = n · (a + b)/(ab) = n · (1/b + 1/a) = n · (a+b)/(ab)

Step 7: Relate to the harmonic mean.

The harmonic mean h of a and b is: h = 2ab/(a+b)

Therefore: (a+b)/(ab) = 2/h

Step 8: Final calculation.

Sum = n · (2/h) = 2n/h

∴ Answer: a

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