# Quantitative Aptitude – Sequence and series

## Slot – 2 – Quantitative Aptitude – Sequence and series – Let a(base1), a(base2)

Q. Let a(base1), a(base2), … be integers such that a(base1)-a(base2)+a(base3)-a(base4)+…+(-1)^(n-1) a(base n)=n, for all n≥1. Then a(base51)+a(base52)+…+a(base1023) equals?
1. 1
2. −1
3. 0
4. 10

Solution: Given , a(base1) – a(base2) + a(base3) – a(base4)+ ……+ (-1)^(n-1) a(base n) = n
Put n = 1, a(base1) = 1
Put n =2 , a(base1) – a(base2) = 2 or 1 – a(base2) = 2
Thus a(base2) = -1
Put n =3, a(base1) – a(base2) + a(base3) = 3, by solving a(base3) = 1
Similarly you will get a(base4) = -1, a(base5) = 1 ..and so on
Thus we can say all odd numbered = 1 all even numbered = -1
And sum of even terms = 0
Sum of odd terms (starting with odd numbered ) = 1
Thus required sum a(base51) + a(base52) + …+ a(base1023) = -1

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