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# CAT 2019 Quant Question with Solution 48

Question:
Let $a_{1}, a_{2}, \ldots$ be integers such that

$a_{1}-a_{2}+a_{3}-a_{4}+\cdots+(-1)^{n-1} a_{n}=n,$ for all $n \geq 1$

Then $a_{51}+a_{52}+\cdots+a_{1023}$ equals

1. -1
2. 10
3. 0
4. 1

Correct Answer: Option: 4

for $n=1, \quad a_{1}=n \Rightarrow a_{1}=1$

for $n=2, \quad a_{1}-a_{2}=2 \Rightarrow a_{2}=-1$

for $n=3, \quad a_{1}-a_{2}+a_{3}=3 \Rightarrow a_{3}=1$

for $n=4, \quad a_{1}-a_{2}+a_{3}-a_{4}=4 \Rightarrow a_{4}=-1$

From the pattern, each odd term = 1 and each even term = -1

$\Rightarrow a_{51}+a_{52}+\cdots+a_{1022}=0$

Therefore the value is equal to $a_{1023}=1$

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