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If the sequence {An } satisfies An = An-1 - An-2, A1 = 0, and A2 = 1,

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If the sequence {An } satisfies An = An-1 - An-2, A1 = 0, and A2 = 1,  [#permalink]

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New post 12 Jul 2018, 00:12
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Difficulty:

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Question Stats:

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[GMAT math practice question]

If the sequence {An} satisfies An = An-1 - An-2, A1 = 0, and A2 = 1, where n is an integer greater than 2, then what is the sum of the first 100 terms of {An}?

A. 1
B. 2
C. 3
D. 4
E. 5

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Re: If the sequence {An } satisfies An = An-1 - An-2, A1 = 0, and A2 = 1,  [#permalink]

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New post 12 Jul 2018, 01:21
1
\(A_ 1 = 0\) and \(A_ 2 = 1\)

\(A_ n = A_{n-1} - A_{n-2}\)

=> \(A_ 3 = A_ 2 - A_ 1 = 1 - 0 = 1\)
=> \(A_ 4 = A_ 3 - A_ 2 = 1 - 1 = 0\)
=> \(A_ 5 = A_ 4 - A_ 3 = 0 - 1 = -1\)
=> \(A_ 6 = A_ 5 - A_ 4 = -1 - 0 = -1\)
=> \(A_ 7 = A_ 6 - A_ 5 = -1 - (-1) = -1 + 1 = 0\)

The sequence repeats 0, 1, 1, 0, -1, -1, 0, 1, 1 .....

The sequence repeats for every 6 terms and the sum of these six terms is 0 + 1 + 1 + 0 + (-1) + (-1) = 0

With in 100, there are 16 such groups till 96th term and the sum of all terms till 96th term is 0

=> \(A_{97} = A_{96} - A_{95} = -1 - (-1) = 0\)

Similarly \(A_{98} = 1\), \(A_{99} = 1\), \(A_{100} = 0\)

Sum of first hundred terms is \(0 + A_{97} + A_{98} + A_{99} + A_{100} = 0 + 0 + 1 + 1 + 0 = 2\)

Hence option B
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Re: If the sequence {An } satisfies An = An-1 - An-2, A1 = 0, and A2 = 1,  [#permalink]

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New post 15 Jul 2018, 17:20
=>

We determine the period of the sequence by examining its terms:
A1 = 0 and A2 = 1.
A3 = A2 – A1 = 1 – 0 = 1
A4 = A3 – A2 = 1 – 1 = 0
A5 = A4 – A3 = 0 – 1 = -1
A6 = A5 – A4 = -1 – 0 = -1
A7 = A6 – A5 = -1 – (-1) = 0
A8 = A7 – A6 = 0 – (-1) = 1
A9 = A8 – A7 = 1 – 0 = 1
A10 = A9 – A8 = 1 – 1 = 0


The sequence has period 6.
The sum of the first six terms is A1 + A2 + … A6 = 0 + 1 + 1 + 0 + (-1) + (-1) = 0.
We apply this fact to determine the sum of the first 100 terms of the sequence:
A1 + A2 + … A100
= ( A1 + A2 + … A6 ) + … + ( A91 + A92 + … A96 ) + A97 + A98 + A99 + A100
= 0 + … + 0 + A97 + A98 + A99 + A100
= A97 + A98 + A99 + A100
= 0 + 1 + 1 + 0 = 2.

Therefore, the answer is B.
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Re: If the sequence {An } satisfies An = An-1 - An-2, A1 = 0, and A2 = 1,  [#permalink]

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New post 19 Jul 2018, 11:43
MathRevolution wrote:
[GMAT math practice question]

If the sequence {An} satisfies An = An-1 - An-2, A1 = 0, and A2 = 1, where n is an integer greater than 2, then what is the sum of the first 100 terms of {An}?

A. 1
B. 2
C. 3
D. 4
E. 5


Let’s list the first few terms to discern a pattern.

A1 = 0
A2 = 1
A3 = 1 - 0 = 1
A4 = 1 - 1 = 0
A5 = 0 - 1 = -1
A6 = -1 - 0 = -1
A7 = -1 - (-1) = 0
A8 = 0 - (-1) = 1
A9 = 1 - 0 = 1

At this point, we can see that the terms repeat themselves in a cycle of 6 numbers: 0, 1, 1, 0, -1, -1 (notice that A7 = A1, A8 = A2, A9 = A3, etc.). Also notice that the sum of the 6 numbers in one cycle is 0. So the sum of all the terms up to and including the 96th term is 0 (notice 96 = 6 x 16). So we really just need to add A97, A98, A99 and A 100. Since A97 = A1 = 0, A98 = A2 = 1, A99 = 1 and A100 = 0, the sum of these 4 terms (and hence the sum of the first 100 terms) is 0 + 1 + 1 + 0 = 2.

Answer: B
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Re: If the sequence {An } satisfies An = An-1 - An-2, A1 = 0, and A2 = 1, &nbs [#permalink] 19 Jul 2018, 11:43
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