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Infinite Sequence The infinite sequence a1, a2, , an, is [#permalink]

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21 Jun 2006, 16:12

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Infinite Sequence
The infinite sequence a1, a2,â€¦, an,â€¦ is such that a1 = 3, a2 = -1, a3 = 6, a4 = -2, and an = an-4 for n > 4. What is the sum of the first 83 terms of the sequence?

Re: Tough Series Problem - Infinite Sequence [#permalink]

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21 Jun 2006, 18:47

C.
a1 = 3
a2 = -1
a3 = 6
a4 = -2
now every term repeats in the pattren as above. so there are 83 = (20x4) + 3 terms.
sum of 83 terms = sum of the first 20x4 terms + sum of the last 3 terms
= 20 (3-1+6-2) + (3-1+6)
= 128

Re: Tough Series Problem - Infinite Sequence [#permalink]

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28 Jun 2006, 14:56

rbcola wrote:

Infinite Sequence The infinite sequence a1, a2,â€¦, an,â€¦ is such that a1 = 3, a2 = -1, a3 = 6, a4 = -2, and an = an-4 for n > 4. What is the sum of the first 83 terms of the sequence?

A. 120 B. 124 C. 128 D. 132 E. 136

This is a pretty straightforward problem once we figure out that an-4 refers to a (n-4 subscript). I took it for a*n -4.

Anyway, its clear that the 1st, 5th, 9th, 13, 17th....terms are identical similarly, the 2nd, 6th, 10th ..... terms are identical.

There are 4 possible values. Now 4 * 20 =80, so for the 1st 80 terms each of the 4 values a1,a2,a3,a4 will be repeated 80/4 = 20 times. Thus sum of the 1st 80 terms = 20 * (3 -1 + 6 -2) = 120. The sum of the 81st, 82nd and 83rd terms is (3 -1 + 6) = 8. Hence the sum is 128.

At first I too took the question as An=An-4, meaning as A5 = A5 - 4.

I thought that I was going mad... I could not figure out the problem, but thanks to you guys, I was able to understand the logic and what it was asking.