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yrozenblum
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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saynchalk
­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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yrozenblum
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

Attachment:
2023-12-01_17-52-49.png
­
First check out these two posts on sequences:
https://anaprep.com/algebra-introducing-sequences/
https://anaprep.com/algebra-a-difficult ... sequences/

We should find out the first few terms of the sequence. We have been given that starting from the third term, each term is the difference of the previous 2 terms. So the terms will be

-1, 1, 2, 1, -1, -2, -1, 1, ...

Each term depends on previous 2 terms. We will get a pattern when a pair of consecutive terms is repeated. We got -1 and 1 and hence now the patterm will start again. 

-1, 1, 2, 1, -1, -2, -1, 1, 2, 1, -1, -2, ...-1, 1, 2, 1, ...

The cyclicity of the pattern is 6. The sum of all these 6 terms is 0. 
So of the 1000 terms, first 996 terms will all add up to 0. Last 4 terms will be -1, 1, 2, 1 which will add up to 3.

Answer (D)­
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­Hi,

I had a doubt here, I followed a different approach and landed up on the naswer to be 0.

My approach:-

Sum = Avg * Number of Terms
Avg = (First Term + Last Term)/2 ; First Term = -1, Last Term = 1 => Avg = 0 Therefore, Sum =0
I understood the approach explained in the above posts but I am confused on how the above method is leading up to a different answer. Appreciate your help!!
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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?

\(a_1=-1\)
\(a_2=1\)
\(a_3=a_2-a_1 = 1 - (-1) = 2\)
\(a_4=a_3-a_2 = 2 - (1) = 1\)
\(a_5=a_4-a_3 = 1 - (2) = -1\)
\(a_6=a_5-a_4 = -1 - (1) = -2\)
\(a_7=a_6-a_5 = -2 - (-1) = -1\)
\(a_8=a_7-a_6 = -1 - (-2) = 1\)

Sequence = {-1,1,2,1,-1,-2,-1,1,2,1,-1,-2,....}
The sum of repeated 6 terms = -1+1+2+1-1-2 = 0

1000 = 6*166 + 4
Sum of first 1,000 terms of the sequence = 0*166 + (-1+1+2+1) = 3

IMO D
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pratiksha1998
­Hi,

I had a doubt here, I followed a different approach and landed up on the naswer to be 0.

My approach:-

Sum = Avg * Number of Terms
Avg = (First Term + Last Term)/2 ; First Term = -1, Last Term = 1 => Avg = 0 Therefore, Sum =0
I understood the approach explained in the above posts but I am confused on how the above method is leading up to a different answer. Appreciate your help!!
­
You are talking about Arithmetic Progression. This is not an AP. The formula you metioned for calculating the average does not work for this.
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