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Two arithmetic progressions {aₙ} and {bₙ} satisfy a₁ = 25, b₁ = 75, and a₁₀₀ + b₁₀₀ = 100. Which of the following statements is necessarily true?

1. The common difference of progression {aₙ} is equal in magnitude but opposite in sign to the common difference of progression {bₙ}.
2. aₙ + bₙ = 100 for every positive integer n.
3. The sequence (a₁ + b₁), (a₂ + b₂), (a₃ + b₃),... is itself an arithmetic progression.
4. Sum of the first 100 terms of (a_r + b_r) equals 10⁴.

Correct answer: aₙ + bₙ = 100 for every positive integer n.

Solution

Let dₐ and d_b be the common differences. Then cₙ = aₙ + bₙ = 100 + (dₐ + d_b)(n-1). Since c₁ = 100 and c₁₀₀ = 100, we get (dₐ + d_b)*99 = 0, so dₐ + d_b = 0, meaning cₙ = 100 for all n.

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