CAT 2019 Shift 2 Paper

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Question : 79 of 100

Marks: +1, -0

Let

a_1,a_2,...

be integers such that

a_1-a_2+a_3-a_4+...+(-1)^{n-1}a_n=n

, for all

n ≥ 1.

Then

a_51+a_52+...+a_1023

equals

-1
0
10
1

Solution:

for

n=1,a_{1}=n ⇒ a_{1}=1

for

n=2, a_{1}-a_{2}=2 ⇒ a_{2}=-1

for

n=3, a_{1}-a_{2}+a_{3}=3 ⇒ a_{3}=1

for

n=4,a_{1}-a_{2}+a_{3}-a_{4}=4 ⇒ a_{4}=-1

From the pattern, each odd term

=1

and each even term

=-1

⇒ a_{51}+a_{52}+⋅s+a_{1022}=0

Therefore the value is equal to

a_{1023}=1

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