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CAT 2019 Slot 2 — QA Question 5

Miscellaneous ProgressionsEasy

Let a1 , a2 be integers such that a1 - a2 + a3 - a4 + ........ + (-1)n-1 an = n , for n ≥ 1. Then a51 + a52 + ........ + a1023 equals

Answer & solution

Correct answer: 1

  • A

    -1

  • 1

  • C

    0

  • D

    10

Solution

Easy

Subtract the relation for n−1n-1 from the one for nn: almost everything cancels, leaving a simple rule for each ana_n. Plug in small nn to spot the pattern 1,−1,1,−1,…1,-1,1,-1,\dots, then count odd and even indices from 5151 to 10231023.

1

Read off the first terms. Use a1−a2+a3−⋯+(−1)n−1an=na_1-a_2+a_3-\dots+(-1)^{n-1}a_n=n for small nn.

n=1: a1=1n=2: a1−a2=2 ⇒ a2=−1n=3: a1−a2+a3=3 ⇒ a3=1\begin{aligned} &n=1:\ a_1=1\\ &n=2:\ a_1-a_2=2\ \Rightarrow\ a_2=-1\\ &n=3:\ a_1-a_2+a_3=3\ \Rightarrow\ a_3=1 \end{aligned}
2

Identify the pattern. Subtracting consecutive relations gives (−1)n−1an=1(-1)^{n-1}a_n=1, so

an=(−1) n−1={+1n odd−1n even\begin{aligned} &a_n=(-1)^{\,n-1}=\begin{cases}+1 & n\ \text{odd}\\ -1 & n\ \text{even}\end{cases} \end{aligned}
3

Count terms from 5151 to 10231023. Total terms =1023−51+1=973=1023-51+1=973. The indices run odd, even, odd, …, starting and ending on odd numbers, so there is one extra odd index.

#odd=487 (each +1),#even=486 (each −1)⇒ ∑k=511023ak=487−486=1\begin{aligned} &\#\text{odd}=487\ (\text{each }+1),\quad \#\text{even}=486\ (\text{each }-1)\\ &\Rightarrow\ \sum_{k=51}^{1023}a_k=487-486=1 \end{aligned}
a51+a52+⋯+a1023=1a_{51}+a_{52}+\dots+a_{1023}=1

Pairing (+1)+(−1)=0(+1)+(-1)=0 wipes out every odd–even pair. With one unpaired odd term left over (since both ends are odd), the sum is just that leftover +1+1.

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CAT 2019 Slot 2 QA Q5: Let a 1 , a 2 be integers such that a 1 - a 2 + a 3 - a 4 + ........ + (-1) n-1 a n = n , for n ≥ 1. Then a — Solution | TheCATExam