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The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2)

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yrozenblum
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The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2) [#permalink] New post   14 Oct 2023, 11:24
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The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_{n−1}-a_{n−2}\) for all integers \(n\geq{3}\). If \(a_1=-1\) and \(a_2=1\), what is the sum of the first 1,000 terms of the sequence?

A. -1
B. 0
C. 2
D. 3
E. 6

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Introvertuprising
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Re: The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2) [#permalink] New post   14 Oct 2023, 12:09
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a1=-1
a2=1
a3=a2-a1=1-(-1) =2
a4=a3-a2=2-1=1
a5=a4-a3=1-2=-1
a6=a5-14=-1-1=-2

a7=a6-a5=-2+1=-1
a8=a7-a6=-1+2=1
a9=a8-a7=1-(-1)=2
the numbers repeat after a6
a1+a2+a3+a4+a5+a6=0
1000/6 =166*6 =996
for the first 966 terms the sum will be zero
997-1000 the sum will be 3
Ans D
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gmatophobia
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Re: The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2) [#permalink] New post   14 Oct 2023, 13:11
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yrozenblum wrote:
The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_{n−1}-a_{n−2}\) for all integers \(n\geq{3}\). If \(a_1=-1\) and \(a_2=1\), what is the sum of the first 1,000 terms of the sequence?

A. -1
B. 0
C. 2
D. 3
E. 6


We can write few terms and find a pattern

\(a_1 = -1\)

\(a_2 = 1\)

\(a_3 = a_2 - a_1 = 1 - (-1) = 2\)

\(a_4 = a_3 - a_2 = 2 - (1) = 1\)

\(a_5 = a_4 - a_3 = 1 - (2) = -1\)

\(a_6 = a_5 - a_4 = -1 - (1) = -2\)

\(a_7 = a_6 - a_5 = -2 - (-1) = -1\)

\(a_8 = a_7 - a_6 = -1 - (2) = 1\)
.
.
.

Therefore we see that the pattern repeats after every 6 terms. The sum of the first 6 terms = 2 + 1- 1 -2 -1 + 1 = 0

For every 6 terms, the sum = 0

1000 = 996 + 4 = 6*(166) + 4

The sum of the first 996 terms = 0

\(997^{\text{th}}\) term = -1

\(998^{\text{th}}\) term = 1

\(999^{\text{th}}\) term = 2

\(1000^{\text{th}}\) term = 1

Sum = -1 + 1 + 2 + 1 = 3

Option D
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Re: The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2) [#permalink] New post   14 Mar 2024, 00:22
gmatophobia wrote:
yrozenblum wrote:
The sequence \(a_1\), \(a_2\), ..., \(a_n\), ... is such that \(a_n=a_{n−1}-a_{n−2}\) for all integers \(n\geq{3}\). If \(a_1=-1\) and \(a_2=1\), what is the sum of the first 1,000 terms of the sequence?

A. -1
B. 0
C. 2
D. 3
E. 6

We can write few terms and find a pattern

\(a_1 = -1\)

\(a_2 = 1\)

\(a_3 = a_2 - a_1 = 1 - (-1) = 2\)

\(a_4 = a_3 - a_2 = 2 - (1) = 1\)

\(a_5 = a_4 - a_3 = 1 - (2) = -1\)

\(a_6 = a_5 - a_4 = -1 - (1) = -2\)

\(a_7 = a_6 - a_5 = -2 - (-1) = -1\)

\(a_8 = a_7 - a_6 = -1 - (2) = 1\)
.
.
.

Therefore we see that the pattern repeats after every 6 terms. The sum of the first 6 terms = 2 + 1- 1 -2 -1 + 1 = 0

For every 6 terms, the sum = 0

1000 = 996 + 4 = 6*(166) + 4

The sum of the first 996 terms = 0

\(997^{\text{th}}\) term = -1

\(998^{\text{th}}\) term = 1

\(999^{\text{th}}\) term = 2

\(1000^{\text{th}}\) term = 1

Sum = -1 + 1 + 2 + 1 = 3

Option D

­Hi, I had a question about getting the final number before 1000 which is a multiple of 6

what I did was I divided 1000 by 6, and got the remainder 4
then I subtracted 4 from 1000 to get 996.

Is this method correct?
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gmatophobia
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Re: The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2) [#permalink] New post   16 Mar 2024, 02:52
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saynchalk wrote:
­Hi, I had a question about getting the final number before 1000 which is a multiple of 6

what I did was I divided 1000 by 6, and got the remainder 4
then I subtracted 4 from 1000 to get 996.

Is this method correct?

Yes, your method is correct ­saynchalk .

We are using the concept 

Dividend = Divisor * Quotient + Remainder­
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Re: The sequence a1, a2, ..., an, ... is such that an = a(n-1) - a(n-2) [#permalink]
16 Mar 2024, 02:52
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