retagged by
439 views
1 votes
1 votes

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. $-200$
retagged by

1 Answer

1 votes
1 votes

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 $:\text{B}$

edited by
Answer:

Related questions

1 votes
1 votes
1 answer
1
soujanyareddy13 asked Jan 20, 2022
411 views
Consider the pair of equations: $x^{2} – xy – x = 22$ and $y^{2} – xy + y = 34.$ If $x>y,$ then $x – y$ equals$7$$8$$6$$4$
1 votes
1 votes
1 answer
2
1 votes
1 votes
1 answer
3
soujanyareddy13 asked Jan 20, 2022
1,284 views
The number of ways of distributing $15$ identical balloons, $6$ identical pencils and $3$ identical erasers among $3$ children, such that each child gets at least four ba...
1 votes
1 votes
1 answer
4
soujanyareddy13 asked Jan 20, 2022
670 views
For a real number $x$ the condition $|3x – 20| + |3x – 40| = 20$ necessarily holds if$9 < x < 14$$6 < x < 11$$7 < x < 12$$10 < x < 15$
1 votes
1 votes
1 answer
5