Question:medium

For a sequence of real numbers \(x_1, x_2, ..., x_n,\) if \(x_1 - x_2 + x_3 - ... + (-1)^{n + 1}x_n =n^2 + 2n\) for all natural numbers n, then the sum \(x_{49} + x_{50}\) equals

Updated On: Jan 15, 2026
  • 2
  • -2
  • 200
  • -200
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The Correct Option is B

Solution and Explanation

The initial term of the sequence is provided.

The relationship between consecutive terms is established as: \(x_1 - x_2 = 8\). Given \(x_1 = 3\), we deduce \(x_2\) by substituting: \(3 - x_2 = 8 \Rightarrow x_2 = -5\).

The next condition for the series is: \(x_1 - x_2 + x_3 = 15\). Substituting the known values \(x_1 = 3\) and \(x_2 = -5\): \(3 - (-5) + x_3 = 15 \Rightarrow 8 + x_3 = 15 \Rightarrow x_3 = 7\).

The pattern of the sequence terms is analyzed: Observation of term values reveals: \(x_1 = 3 = (+1)(2×1+1)\) \(x_2 = -5 = (-1)(2×2+1)\) \(x_3 = 7 = (+1)(2×3+1)\)

Based on this analysis, the general term of the sequence is determined to be: \(x_n = (-1)^{n+1}(2n+1)\).

The values for specific terms are computed: \(x_{49} = (-1)^{50}(2×49+1) = (-1)^{50}(99) = +99\) \(x_{50} = (-1)^{51}(2×50+1) = (-1)^{51}(101) = -101\)

The sum of these two terms is calculated: \(x_{49} + x_{50} = 99 + (-101) = -2\).

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