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M08-08

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M08-08  [#permalink]

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New post 15 Sep 2014, 23:36
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A sequence is defined as follows:
\(A_1 = 1\)
\(A_2 = -1\)
\(A_{n+1} = A_n + 2A_{n-1}\)

What is the sum \(A_1 + A_2 + ... + A_{1001}\)?


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

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Re M08-08  [#permalink]

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New post 15 Sep 2014, 23:36
Official Solution:


A sequence is defined as follows:
\(A_1 = 1\)
\(A_2 = -1\)
\(A_{n+1} = A_n + 2A_{n-1}\)

What is the sum \(A_1 + A_2 + ... + A_{1001}\)?


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

Write out more terms of this sequence: \(A_3 = 1\); \(A_4 = -1\); \(A_5 = 1\). Even terms equal -1 and odd terms equal 1. Therefore, the sum \(A_1 + A_2 + ... + A_{1001} = 1\)

Answer: D
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Re: M08-08  [#permalink]

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New post 19 Jul 2016, 17:49
How did you get A3 = 1 , can you please show the setup ?
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New post 20 Jul 2016, 01:14
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Re M08-08  [#permalink]

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New post 29 Jul 2016, 09:46
I think this is a high-quality question and I agree with explanation.
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Re M08-08  [#permalink]

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New post 20 May 2017, 08:14
I think this is a high-quality question and I agree with explanation. This is surely a 600 Lv question
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Re: M08-08  [#permalink]

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New post 26 Jun 2017, 15:10
I am still confused after reading your explanation wouldn't the answer be -1?
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Re: M08-08  [#permalink]

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New post 26 Jun 2017, 20:16
POQBigwill wrote:
I am still confused after reading your explanation wouldn't the answer be -1?


Why do you think the answer would be -1? Hod did you get this?

The solutions says that even terms equal -1 and odd terms equal 1. We have the sum of odd number of terms (1001). The sum would be the same for any other odd number of terms: 1 + (-1) + 1 = 1.
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Re: M08-08  [#permalink]

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New post 19 Dec 2018, 21:26
Dear Experts chetan2u Bunuel

Even I got the answer -1 (I am sure I used the wrong formula and thinking)

When does the below formula hold true?
Sum of an infinite geometric progression with common ratio <1 is b/1-r (where b is the first term)

So (I may have wrongly assumed common ratio =-1-1 =-2 which is <1
hence according to the above formula
b/1-r = 1/(1-2) = -1
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Re: M08-08  [#permalink]

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New post 19 Dec 2018, 22:20
deddex wrote:
Dear Experts chetan2u Bunuel

Even I got the answer -1 (I am sure I used the wrong formula and thinking)

When does the below formula hold true?
Sum of an infinite geometric progression with common ratio <1 is b/1-r (where b is the first term)

So (I may have wrongly assumed common ratio =-1-1 =-2 which is <1
hence according to the above formula
b/1-r = 1/(1-2) = -1


The sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.

Take a look at the highlighted words.

1. We are not asked to find the sum of the infinite sum, we are asked to find the sum of the first 1,001 terms.
2. The sequence is: 1, -1, 1, -1, 1, .... While this IS in fact a geometric progression, the common ratio is -1 (you can find it by dividing any two consecutive terms). As you can see the absolute value of the ratio is not less than 1.

For more on sequences check below:

12. Sequences



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Hope it helps.
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Collection of Questions:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M08-08 &nbs [#permalink] 19 Dec 2018, 22:20
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