# CAT 2021 Set-2 | Quantitative Aptitude | Question: 5

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For a sequence of real numbers $x_{1}, x_{2}, \dots , x_{n},$ if $x_{1} – x_{2} + x_{3} – \dots + (-1)^{n+1} x_{n} = n^{2} + 2n$ for all natural numbers $n,$ then the sum $x_{49} + x_{50}$ equals

1. $2$
2. $-2$
3. $200$
4. $-200$

## 1 Answer

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

And, $x_{1} – x_{2} + x_{3} –\dots + (-1)^{n+1}x_{n} = n^{2}+2n\quad \longrightarrow(1)$

Now, put the $n=49 \; \text{and} \; n=50,$ we get

• $x_{1} – x_{2} + x_{3} –\dots +x_{49} = 49^{2}+2 \times 49\quad \longrightarrow(2)$

• $x_{1} – x_{2} + x_{3} –\dots +x_{49} – x_{50} = 50^{2}+2 \times 50\quad \longrightarrow(3)$

Subtract equation $(3)$ from equation $(2).$

$x_{1} – x_{2} + x_{3} –\dots +x_{49} – (x_{1} – x_{2} + x_{3} –\dots +x_{49} – x_{50}) = 49^{2}+2 \times 49 – (50^{2}+2 \times 50)$

$\Rightarrow x_{50} = 49^{2} – 50^{2} + 2 \times 49 – 2 \times 50$

$\Rightarrow x_{50} = (49+50)(49–50) + 2 (-1) \quad {\color{Green} {[\because a^{2} – b^{2} = (a-b)(a+b)]}}$

$\Rightarrow x_{50} = -99-2$

$\Rightarrow \boxed{x_{50} = -101}$

Now, put $n=48,$ in the equation $(1).$

$x_{1} – x_{2} + x_{3} –\dots +x_{47} – x_{48} = 48^{2}+2 \times 48\quad \longrightarrow(4)$

Subtract equation $(4)$ from equation $(2).$

$x_{1} – x_{2} + x_{3} –\dots -x_{48} +x_{49} -(x_{1} – x_{2} + x_{3} –\dots -x_{48}) = 49^{2}+2 \times 49–48^{2} –2 \times 48$

$\Rightarrow x_{49} = 49^{2} – 48^{2} + 2 \times 49 – 2 \times 48$

$\Rightarrow x_{49} = (49+48)(49-48)+2(49-48)$

$\Rightarrow x_{49} = 97+2$

$\Rightarrow \boxed{x_{49} = 99}$

$\therefore$ The sum $\; x_{49}+x_{50} = 99-101 = -2.$

Correct Answer $:\text{B}$

Answer:

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