CAT 2021 Set-3 | Quantitative Aptitude | Question: 1

Question

Consider a sequence of real numbers $x_{1}, x_{2}, x_{3}, \dots  $ such that $x_{n+1} = x_{n} + n – 1$ for all $n \geq 1.$ If $x_{1} = -1$ then $x_{100}$ is equal to 

• A. $4950$
• B. $4850$
• C. $4849$
• D. $4949$

Answer 1

Given that, $x_{1}, x_{2}, x_{3}, \dots, x_{n},  x_{n+1} \in \mathbb{R}$

And $x_{n+1} = x_{n}+n-1 \;; \forall n \geq 1$

Now,
$\qquad \begin{array}{} x_{1} = -1 \\ x_{2} = x_{1} +0 \\ x_{3} = x_{2} +1 \\ x_{4} = x_{3} +2 \\  x_{5} = x_{4} +3 \\ x_{6} = x_{5} +4 \\ \;\vdots \quad \vdots \quad  \vdots \quad  \vdots  \;\;  \vdots  \\ x_{100} = x_{99} +98 \\\hline \end{array}$

Adding all the terms,

$x_{1}+x_{2}+x_{3}+ \dots +x_{100} = x_{1}+x_{2}+x_{3}+ \dots +x_{99}+(-1)+0+1+2+3+\dots +98$

$\Rightarrow x_{100}= 1+2+3+\dots +98-1$

$\Rightarrow x_{100}= \underbrace{\frac{98(98+1)}{2}}_{\text{Sum of the first 98 numbers}}-1$

$\Rightarrow x_{100} = 49 \times 99-1$

$\Rightarrow x_{100} = 4851-1$

$\Rightarrow \boxed{x_{100} = 4850}$

Correct Answer $:\text{B}$