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

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

Let a(base1), a(base2) – Video Solution

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 nâ‰¥1. Then a(base51)+a(base52)+…+a(base1023) equals?

1

âˆ’1

0

10

Answer: 1

Solution:
Given , a(base1) – a(base2) + a(base3) â€“ a(base4)+ â€¦â€¦+ (-1)^(n-1) a(base n) = n
Put n = 1, a(base1) = 1
Put n =2 , a(base1) – a(base2) = 2 or 1 â€“ 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
We are now asked to find out
a(base51) + a(base52) + â€¦+ a(base1023) = 1 + (-1) + 1 + (-1)…
Starting with a(51), if the total number of terms is even: all terms will cancel out and the result will be 0.
So,
a(51) + a(52) = 0
a(51) + a(52) + a(53) + a(54) = 0
a(51) + a(52) + a(53) + a(54) … a(1022)= 0
a(51) + a(52) + a(53) + a(54) … a(1022) + a(1023)= 0 + a(1023) = 0 + 1 = 1
So, our answer for this sum would be 1.

a) 1000+ Videos covering entire CAT syllabus b) 2 Live Classes (online) every week for doubt clarification c) Study Material & PDFs for practice and understanding d) 10 Mock Tests in the latest pattern e) Previous Year Questions solved on video