soujanyareddy13
asked
in Quantitative Aptitude
Jan 20, 2022
retagged
Mar 23, 2022
by Lakshman Bhaiya

319 views
1 vote

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 $:\text{B}$

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 $:\text{B}$