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# The infinite sequence a1, a2,, an, is such that a1 = 3, a2 =

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The infinite sequence a1, a2,, an, is such that a1 = 3, a2 = [#permalink]

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24 Aug 2008, 12:53
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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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24 Aug 2008, 15:15
sarzan wrote:
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

C128.

For every 4 terms first 4 values will repeat.

Sum of first 4 values = 3-1+6-2=6

Sum of 83 terms= sum of 80 terms + last 3 terms(equal to first 3 terms)
= (80/4)*6 + (3-1+6)
= 120+8=128
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24 Aug 2008, 15:30
sarzan wrote:
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

C

83/4 = 20 R 3

This means that there are 20 sets of 3+(-1)+6+(-2).
Sum of first 80 terms = 20*(3-1+6-2) = 20*6 = 120

The remaining three terms are 3, -1, and 6, since the sequence is repeating
Sum of 3 terms = 3-1+6 = 8

Sum of 83 terms = 128

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24 Aug 2008, 16:05
but why will the 4 terms repeat themselves? What would terms 5 through 8 be like?

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24 Aug 2008, 16:22
sarzan wrote:
but why will the 4 terms repeat themselves? What would terms 5 through 8 be like?

an=an-4
a5=a(5-1)=a1
a6=a2
a7=a3
a8=a4

a9=a5=a1
.
.
.
so on.
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24 Aug 2008, 20:42
x2suresh wrote:
sarzan wrote:
but why will the 4 terms repeat themselves? What would terms 5 through 8 be like?

an=an-4
a5=a(5-1)=a1
a6=a2
a7=a3
a8=a4

a9=a5=a1
.
.
.
so on.

mind explaining where you came up with that ? Im having trouble seeing how to use an = an-4 for when n=5, for example

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24 Aug 2008, 22:11
pmenon wrote:
x2suresh wrote:
sarzan wrote:
but why will the 4 terms repeat themselves? What would terms 5 through 8 be like?

an=an-4
a5=a(5-1)=a1
a6=a2
a7=a3
a8=a4

a9=a5=a1
.
.
.
so on.

mind explaining where you came up with that ? Im having trouble seeing how to use an = an-4 for when n=5, for example

$$a_n=a_{n-4}$$
$$a_5=a_{5-1}=a_1$$
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24 Aug 2008, 22:17
Quote:
$$a_n=a_{n-4}$$
$$a_5=a_{5-1}=a_1$$

the question says that any nth term for n>4, that term = a(n-4)
so a5 = a(5-4) = a1

similarly a6 = a(6-4) = a2
so on...............

so now after a4, it is simply a repetation of first 4 terms...
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Re: PS - sequence   [#permalink] 24 Aug 2008, 22:17
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