CAT 2021 Question Paper | Quant Slot 3

CAT 2021 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 16 : Consider a sequence of real numbers \(x_{1}, x_{2}, x_{3}, \ldots\) such that \(x_{n+1}=x_{n}+n-1\) for all \(n \geq 1 .\) If \(x_{1}=-1\) then \(x_{100}\) is equal to

Video Explanation

Explanatory Answer

x_n+1 = x_n + n - 1
x₁ = -1
x₂ = x₁ + (1 - 1)
x₃ = x₂ + (2 - 1)
x₃ = x₁ + (1 - 1) + (2 - 1)
x₄ = x₃ + (3 - 1)
x₄ = x₁ + (1 - 1) + (2 - 1) + (3 - 1)
Similarly we see that,
x₁₀₀ = x₁ + (1 - 1) + (2 - 1) + (3 - 1) + … + (99 - 1)
x₁₀₀ = x₁ + (1 + 2 + 3 + 4 + … + 98 + 99) - 99 (1)
x₁₀₀ = x₁ + (1 + 2 + 3 + 4 + … + 98)
x₁₀₀ = (-1) + (\( \frac { 98 \times 99 } { 2 } \))
x₁₀₀ = (-1) + 4851
x₁₀₀ = 4850

The question is " Consider a sequence of real numbers \(x_{1}, x_{2}, x_{3}, \ldots\) such that \(x_{n+1}=x_{n}+n-1\) for all \(n \geq 1 .\) If \(x_{1}=-1\) then \(x_{100}\) is equal to "

Hence, the answer is '4850'

Choice C is the correct answer.

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