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The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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30 Aug 2010, 10:26
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The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_1 = 2\), \(a_2 = 3\), \(a_3 = 5\), \(a_4 = 1\), and \(a_n = a_{n4}\) for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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udaymathapati wrote: The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = 3, a3 = 5, a4 = 1, and an = an4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80 Given: \(a_1=2\), \(a_2=3\), \(a_3=5\) and \(a_4=1\). Also \(a_n=a_{n4}\), for \(n>4\). Since \(a_n=a_{n4}\) then: \(a_5=a_1=2\); \(a_6=a_2=3\); \(a_7=a_3=5\); \(a_8=a_4=1\); \(a_9=a_5=a_1=2\); and so on. So we have groups of 4: {2, 3, 5, 1}, the sum of each of such group is \(23+51=3\). 97 terms consist of 24 full groups plus \(a_{97}=a_1=2\), so the sum of first 97 terms of the sequence is \(24*3+2=74\). Answer: B.
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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25 Oct 2010, 04:01
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the sequence is repeating after each 4 terms.
i.e 2,3,5,1 ,2,3,5,1 ,2,3,5,1 ,2,3,5,1 ....till 97th term.
Now (2,3,5,1 ),(2,3,5,1) ,(2,3,5,1 ),......................(2,3,5,1) ,1
97 = 24 *4 + 1
Means 24 similar like (2,3,5,1 ) and one more that is first term of the series
the sum of the first four term is = 3
Hence sum is 24*3 + 2 = 74
Hence answer is B.
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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ashiima wrote: The infinite sequence a1, a2,�, an,� is such that a1 = 2, a2 = 3, a3 = 5, a4 = 1, and an = an4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80 Given: a(n) = a(n4) i.e. the nth term is same as (n4)th term e.g. 5th term is same as 1st term. 6th term is same as 2nd term etc The sequence becomes: 2, 3, 5, 1, 2, 3, 5, 1, 2, 3, 5, 1 ... The sum of each group of 4 terms = 23+51 = 3 How many such groups of 4 are there in the first 97 terms? When you divide by 4, you get 24 as quotient and 1 as remainder. This means you get 24 complete groups of 4 terms each and 1 extra term i.e. the 97th term. Sum of 24 groups of 4 terms each = 24*3 = 72 The 97th term will be 2 since it is the first term of the next group of four terms. Sum of first 97 terms = 72+2 = 74
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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03 Jan 2012, 11:37
The sum of each set of 4 is 3 so to find the answer find the floor of 97/4 (this is 24) this leaves one term unaccounted for so the equation becomes 24*3+2=74.



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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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04 Jan 2012, 07:13
can you explain a bit more. from a5 to a93 i get an AP for which the sum is = 4371 from a1 to a4, the sum = 3 Total sum= 4374. what am i doing wrong?



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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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04 Jan 2012, 11:05
Baten80 wrote: The infinite sequence a1, a2,, an, is such that a1 = 2, a2 = 3, a3 = 5, a4 = 1, and a_n = a_n4 for n > 4. What is the sum of the first 97 terms of the sequence? A. 72 B. 74 C. 75 D. 78 E. 80 Given for n>4 => a_n = a_n4 => a_5 = a_1 & a_9 = a_5 = a_1 so taking the above facts => (a_1+a_2+a_3+a_4+a_1+.....) => 24(a_1+a_2+a_3+a_4)+a_1 (since it is the repitition of the 4 terms so 24*4 = 96) so adding the above we get => 74 (B)



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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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04 Jan 2012, 11:26
Nice one, a pattern problem with an addition twist. a1 = 2, a2 = 3, a3 = 5, a4 = 1 What an = a(n4) means that from a5 onwards, every number in the sequence will be equal to 4 numbers preceding that number. Therefore, a5 = a1 = 2, a6 = a2 = 3,... Notice that this is a 4element set pattern. If we sum a1a4, we get 3. What is the closet multiple of 4 to 97  96! So, there are 24 such complete sets. Added to that is 97th number, which is first in the sequence = 2. So, we get (24 x 3) + 2 = 74 > B
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Sum of a sequence [#permalink]
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15 Jul 2014, 20:43
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\(a_1 = 2\) \(a_2 = 3\) \(a_3 = 5\) \(a_4 = 1\) \(a_5 = a_{(54)} = a_1 = 2\) \(a_6 = a_{(64)} = a_2 = 3\) \((a_1 + a_2 + a_3 + a_4) = (a_5 + a_6 + a_7 + a_8) = (a_9 + a_{10} .....)\) and so on Addition upto \(a_{96}\) is repeated 24 times \((a_1 + a_2 + a_3 + a_4) = 2  3 + 5  1 = 3\) \(a_1 + a_2 + a_3 + ........ + a_{96} = 3 * 24 = 72\) \((a_1+a_2+a_3+ ........ + a_{96}) + a_{97} = 72 + 2 = 74\) Answer = B
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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28 Feb 2017, 02:54
From the equation, a5 = a1, a6 = a2, a7 = a3, a8 = a4 and again a9 =a5 =a1 So sum of 97 terms =sum of 96 terms +97th term = 24*( a1 +a2 +a3 +a4) + a1 = 24*(2 3 +5 1) + 2 = 72 +2 = 74. Option B



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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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udaymathapati wrote: The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_1 = 2\), \(a_2 = 3\), \(a_3 = 5\), \(a_4 = 1\), and \(a_n = a_{n4}\) for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80 We are given that a(1) = 2, a(2) = 3, a(3) = 5, and a(4) = 1. Since a(n) = a(n4), we see that: a(5) = a(54) = a(1) = 2, a(6) = a(64) = a(2) = 3, a(7) = a(74) = a(3) = 5, and a(8) = a(84) = a(4) = 1 As we can see, the terms repeat themselves in a cycle of 4. Since the sum of the terms a(1) to a(4) inclusive is [2 + (3) + 5 + (1)] = 3, the sum of a(5) to a(8) will also be 3, as will the sum of a(9) to a(12), etc. We see from each “grouping” of 4 numbers, we have a sum of 3, and from 1 to 96 inclusive there are: (96  4)/4 + 1 = 92/4 + 1 = 24 such groupings. Thus, the sum of those 24 groupings is 24 x 3 = 72. We must also add the value of a(97), which equals 2. Thus, the sum of the first 97 terms is: 24 x 3 + 2 = 72 + 2 = 74. Answer: B
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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09 Oct 2017, 12:33
udaymathapati wrote: The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_1 = 2\), \(a_2 = 3\), \(a_3 = 5\), \(a_4 = 1\), and \(a_n = a_{n4}\) for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80 "\(a_n = a_{n4}\) for n > 4." < what is the point of this information? You could still solve for the problem without it, so what was the test makers intention on giving us it? Anyway, Value of n = 1,2,3,4  Value of An= (2) + (3) + (5) + (1) = 3; and for A5, A6, A7,A8, given n>4 = n4, value for n again equals 1,2,3,4  Value for An = 2) + (3) + (5) + (1) = 3 Therefore, the sum of the first 8 digits of 97 equals 6 97/8 = 12; 12 x 6 = 72 since the next digit in line is the start of a new cycle (i.e. equal to A1), that digits value is 2 thus 72 + 2 = 74 Answer (B) But really, other than determining the next four digits, I do not know the purpose of the above piece of information; seeing as An4 for the ninth digit would make A5, which no value was given, it was confusing at first  does the value of An, where n>=9, would repeat or not. I mean, giving the test structure, and cyclicality, I knew that I could stop at 8 and solve for it. However, it just seems weird since what is given really doesn't account for the value of A94 (A5) since they never give us the value of A5, you have to assume it repeats. It seems clearer to state this the question without An. But, I still solved it because I knew what they were looking for for this category of question.



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The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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28 Feb 2018, 10:14
udaymathapati wrote: The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_1 = 2\), \(a_2 = 3\), \(a_3 = 5\), \(a_4 = 1\), and \(a_n = a_{n4}\) for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80 Can anyone provide me with a quality information(website, youtube etc) on GMAT sequences. I cant seem to decode such sort of questions



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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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28 Feb 2018, 10:18
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dave13 wrote: udaymathapati wrote: The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_1 = 2\), \(a_2 = 3\), \(a_3 = 5\), \(a_4 = 1\), and \(a_n = a_{n4}\) for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80 Can anyone provide me with a quality information(website, youtube etc) on GMAT sequences. I cant seem to decode such sort of questions 12. Sequences For ALL other subjects check Ultimate GMAT Quantitative Megathread (I pointed you to this topic multiple times) Hope it helps.
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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28 Feb 2018, 12:50
ScottTargetTestPrep wrote: udaymathapati wrote: The infinite sequence \(a_1\), \(a_2\),…, \(a_n\), … is such that \(a_1 = 2\), \(a_2 = 3\), \(a_3 = 5\), \(a_4 = 1\), and \(a_n = a_{n4}\) for n > 4. What is the sum of the first 97 terms of the sequence?
A. 72 B. 74 C. 75 D. 78 E. 80 We are given that a(1) = 2, a(2) = 3, a(3) = 5, and a(4) = 1. Since a(n) = a(n4), we see that: a(5) = a(54) = a(1) = 2, a(6) = a(64) = a(2) = 3, a(7) = a(74) = a(3) = 5, and a(8) = a(84) = a(4) = 1 As we can see, the terms repeat themselves in a cycle of 4. Since the sum of the terms a(1) to a(4) inclusive is [2 + (3) + 5 + (1)] = 3, the sum of a(5) to a(8) will also be 3, as will the sum of a(9) to a(12), etc. We see from each “grouping” of 4 numbers, we have a sum of 3, and from 1 to 96 inclusive there are: (96  4)/4 + 1 = 92/4 + 1 = 24 such groupings. Thus, the sum of those 24 groupings is 24 x 3 = 72. We must also add the value of a(97), which equals 2. Thus, the sum of the first 97 terms is: 24 x 3 + 2 = 72 + 2 = 74. Answer: B Hello Sir Why here are you deducting 4 from 96 in the numerator (96  4)/4 + 1 = 92/4 + 1 = 24 why didnt you write 96 /4 + 1 Another question why did you take initially take 96 to calculate the sum and than added value of 97 .. why not to take 97 intially i would appreaciate explanation thank you !




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