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Infinite Sequence

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by dtweah » Thu May 28, 2009 6:03 am
The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and a_n = a_n-4 for n > 4. What is the sum of the first 97 terms
of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80
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by PAB2706 » Thu May 28, 2009 6:37 am
IMO B

observation:=

after every four itteration the values repeat..

the total of a1+a2+a3+a4 = 3

97/4=24 and remainder 1

thus 24x3 + a1 ie 72+2=74
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by Thouraya » Wed Mar 02, 2011 12:22 am
Can you please explain this further? How do you know that the values repeat? I tried finding the difference, but there's no consistency or pattern..
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by stormier » Wed Mar 02, 2011 10:39 am
Thouraya wrote:Can you please explain this further? How do you know that the values repeat? I tried finding the difference, but there's no consistency or pattern..
an = a(n-4)

=> a5 = a(5-4) = a1

a9=a(9-4)=a5=a1

similarly;

a1=a5=a9=a13=....a93=a97 = 2
a2=a6=a10=.....=a94 = -3
a3=a7=a11=... =a95 = 5
a4=a8=a12=.......=a96 = -1

sum = 24x(2-3+5-1) + 2 = 74
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by Anurag@Gurome » Wed Mar 02, 2011 7:28 pm
Solution:

The sequence is 2, -3, 5, -1, 2, -3, 5, -1......
So, a1 = 2 = a5 = a9 = a(4k+1), where k is any integer >= 0.
Similarly a2 = -3 = a6 = a10 = a(4k+2)
a3 = 5 = a7 = a11 = a(4k+3)
a4 = -1 = a8 = a12 = a(4k+4)

Note that the sum of first 4 terms is 3 = sum of next 4 terms and so on.
So, upto 96, the sum of all terms is (96/4)*3 = 72.
Since 97 = 4*24+1, the 97th term is 2.
Hence, the sum of 97 terms is 72+2 = 74.
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