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CAT 2021 Slot 2 — QA Question 13

Miscellaneous ProgressionsEasy

For a sequence of real numbers x1, x2, …, xn, if x1 - x2 + x3 - … + (-1)(n+1) xn = n2 + 2n for all natural numbers n, then the sum x49 + x50 equals.

Answer & solution

  • -2

  • B

    2

  • C

    -200

  • D

    200

Solution

Easy

The defining relation holds for every nn. Subtract the relation for n−1n-1 from the relation for nn: this isolates the single term (−1)n+1xn(-1)^{n+1}x_n and gives a closed form for xnx_n. Then read off x49x_{49} and x50x_{50}.

1

Subtract consecutive relations. Let Sn=x1−x2+⋯+(−1)n+1xn=n2+2nS_n=x_1-x_2+\cdots+(-1)^{n+1}x_n=n^{2}+2n. Then Sn−Sn−1=(−1)n+1xnS_n-S_{n-1}=(-1)^{n+1}x_n.

(−1)n+1xn=Sn−Sn−1⇒ (−1)n+1xn=(n2+2n)−((n−1)2+2(n−1))⇒ (−1)n+1xn=2n+1(expand and cancel)\begin{aligned} &(-1)^{n+1}x_n=S_n-S_{n-1}\\ &\Rightarrow\ (-1)^{n+1}x_n=\big(n^{2}+2n\big)-\big((n-1)^{2}+2(n-1)\big)\\ &\Rightarrow\ (-1)^{n+1}x_n=2n+1 \quad\text{(expand and cancel)} \end{aligned}
2

Closed form for xnx_n. Divide by (−1)n+1(-1)^{n+1}.

xn=(−1)n+1(2n+1)\begin{aligned} &x_n=(-1)^{n+1}(2n+1) \end{aligned}

Check n=1n=1: x1=3x_1=3 (matches S1=3S_1=3); n=2n=2: x2=−5x_2=-5 (so x1−x2=8=S2x_1-x_2=8=S_2). Good.

3

Evaluate the required terms. Use the formula at n=49,50n=49,50.

x49=(−1)50(99)=99x50=(−1)51(101)=−101⇒ x49+x50=99−101=−2\begin{aligned} &x_{49}=(-1)^{50}(99)=99\\ &x_{50}=(-1)^{51}(101)=-101\\ &\Rightarrow\ x_{49}+x_{50}=99-101=-2 \end{aligned}
x49+x50=−2x_{49}+x_{50}=-2
CAT 2021 Slot 2 QA Q13: For a sequence of real numbers x 1 , x 2 , …, x n , if x 1 - x 2 + x 3 - … + (-1) (n+1) x n = n — Solution | TheCATExam