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Infinite Sequence The infinite sequence a1, a2, , an, is

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Infinite Sequence The infinite sequence a1, a2, , an, is [#permalink] New post 21 Jun 2006, 16:12
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Infinite Sequence
The infinite sequence a1, a2,…, an,… 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
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 [#permalink] New post 21 Jun 2006, 16:25
C. 128

Given an = an-4 (where n,n-4 is subscript)

i.e. a5 = a1, a6 = a2, a7= a3, a8 = a4 and so on

continuing...

a77 = a1, a28 = a2, a29 = a3, a80 = a4
a81 = a1, a82 = a2, a83 = a3

Sum (a1+a2+a3+a4) = 6

Sum of 83 terms = 80x6/4 + (3-1+6) = 128
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Re: Tough Series Problem - Infinite Sequence [#permalink] New post 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
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 [#permalink] New post 22 Jun 2006, 03:25
128.

(3*21) + (-1*21) + (6*21) + (-1*20)
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Re: Tough Series Problem - Infinite Sequence [#permalink] New post 28 Jun 2006, 14:56
rbcola wrote:
Infinite Sequence
The infinite sequence a1, a2,…, an,… 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.



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 [#permalink] New post 28 Jun 2006, 15:12
Agree with C...

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.
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 [#permalink] New post 29 Jun 2006, 01:29
Given An = A(n-4)

Hence A5 = A1, A6 = A2, A7=A3, A8=A4

A1+A2+A3+A4.............+A80 = 6*(20) = 120

A81 = A1, A82=A2, A83=A3

Hence 120 + 8 = 128
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 [#permalink] New post 29 Jun 2006, 06:32
This was not that tough.

I got 128.

a1 = 3
a2= -1
a3 = 6
a4 = -2
a5 = 3
and so on....

Sum of first 80 terms = 20 (3-1+6-2) = 120
Sum of first 83 terms = 120 +3-1+6 = 128
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  [#permalink] 29 Jun 2006, 06:32
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