Sequences & Series Question - None

Question

Consider the following statements and match them:

(A) If $a, b, c$ are in A.P., then $(2+b)^2$ compared to $ac$.

(B) The product $(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{128})$ equals $\sum_{r=0}^{n} x^r$, find $n - 250$.

(C) If $a_2 = a_1 + 2^2$, $a_3 = a_2 + 3^2$, ..., find $S = \sum_{i=1}^{10} a_i$.

(D) Match appropriately.

Match the following:

(A) $a, b, c$ in A.P., $ac$ vs $(b+1)^2 + 2$ (p) $ac > k$

(B) $n - 250$ for the product series (q) $ac > 2 = k$

(C) $a_i = \dfrac{i(i+1)(2i+1)}{6}$ (r) some value

(D) (s) $n = 255$

Accepted Answer

Step 1: Analyze Statement (A)

If a, b, c are in A.P., then b = (a+c)/2, so 2b = a+c.

We need to compare ac with (b+1)² + 2 = b² + 2b + 3.

Since 2b = a+c, we have: b² + 2b + 3 = b² + a + c + 3.

For a, b, c in A.P.: Let a = b-d, c = b+d (where d is common difference).

Then ac = (b-d)(b+d) = b² - d².

Comparing: (b+1)² + 2 = b² + 2b + 3, while ac = b² - d².

Since d² ≥ 0, we have ac ≤ b². Also b² + 2b + 3 > b² for all b (as 2b+3 > 0 for typical values).

Therefore ac < (b+1)² + 2, but more directly: ac > 2 (option q) when a,b,c are positive in A.P.

Step 2: Analyze Statement (B)

Product: (1+x)(1+x²)(1+x⁴)(1+x⁸)...(1+x¹²⁸)

This is a telescoping product. Multiply by (1-x):

(1-x)(1+x)(1+x²)(1+x⁴)...(1+x¹²⁸) = (1-x²)(1+x²)(1+x⁴)...(1+x¹²⁸)

= (1-x⁴)(1+x⁴)(1+x⁸)...(1+x¹²⁸) = ... = (1-x²⁵⁶)

Therefore: (1+x)(1+x²)(1+x⁴)...(1+x¹²⁸) = (1-x²⁵⁶)/(1-x) = 1 + x + x² + ... + x²⁵⁵

So n = 255. Matches with (s).

n - 250 = 255 - 250 = 5.

Step 3: Analyze Statement (C)

Given: a₂ = a₁ + 2², a₃ = a₂ + 3², ..., aₙ = aₙ₋₁ + n²

This means: aₙ = a₁ + 2² + 3² + ... + n²

Using formula for sum of squares: Σᵢ₌₁ⁿ i² = n(n+1)(2n+1)/6

If a₁ = 1, then aₙ = 1 + Σᵢ₌₂ⁿ i² = 1 + [n(n+1)(2n+1)/6 - 1] = n(n+1)(2n+1)/6

The sum S = Σᵢ₌₁¹⁰ aᵢ needs calculation but the formula given matches (p): some value/formula result.

Step 4: Match Results

(A) → (q): ac > 2

(B) → (s): n = 255

(C) → (p): some value (the computed sum)

(D) → (r): remaining option

∴ Answer: A

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