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

Question

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 2023-12-01_17-52-49.png

gmatophobia · Accepted Answer

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^{	ext{th}}  term = -1

998^{	ext{th}}  term = 1

999^{	ext{th}}  term = 2

1000^{	ext{th}}  term = 1

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

Option D

Introvertuprising · Answer

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

saynchalk · Answer

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?

gmatophobia · Answer

Yes, your method is correct  saynchalk  .

We are using the concept

Dividend = Divisor * Quotient + Remainder

KarishmaB · Answer

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)

pratiksha1998 · Answer

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!!

Kinshook · Answer

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

KarishmaB · Answer

You are talking about Arithmetic Progression. This is not an AP. The formula you metioned for calculating the average does not work for this.

findingmyself · Answer

Another approach: Please rectify

A1=-1, A2=1, A3=A2-A1, A4=A3-A2............A1000=A999-A998, A999=A998-A997

Summing all of this: A1+A2-A1+A999
----> A999=A3 (Since pattern is repeating and A3*333=A999)
New Sum= A1+A2-A1+A3= -1+1+1+2=3

findingmyself · Answer

Another approach: Please rectify

A1=-1, A2=1, A3=A2-A1, A4=A3-A2............A1000=A999-A998, A999=A998-A997

Summing all of this: A1+A2-A1+A999 
 ----> A999=A3 (Since pattern is repeating and A3*333=A999) 
 New Sum= A1+A2-A1+A3= -1+1+1+2=3

Pr4n · Answer

Even though I got it correct, but to see the pattern we need to calculate till a8 minimum to be sure it is repeating, which takes significant time in testing conditions. Is there a even faster way, for example i took 4min to do the question in official mock. Maybe test anxiety also had a role in that but let me know if someone has a even better way to do this.

AditiDeokar · Answer

+1 same doubt. Would the give-away to look for a pattern here be the fact that the answer choices are all in single digits?

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

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

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