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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 \(a_n = a_{n-4}\) for n > 4. What is the sum of the first 97 terms of the sequence?

The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 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 \(a_n=a_{n-4}\), for \(n>4\).

Since \(a_n=a_{n-4}\) 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 \(2-3+5-1=3\). 97 terms consist of 24 full groups plus \(a_{97}=a_1=2\), so the sum of first 97 terms of the sequence is \(24*3+2=74\).

The infinite sequence a1, a2,�, an,� is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 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(n-4) i.e. the nth term is same as (n-4)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 = 2-3+5-1 = 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.

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_n-4 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_n-4 => 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)

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(n-4) 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 4-element set pattern. If we sum a1-a4, 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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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 \(a_n = a_{n-4}\) 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(n-4), we see that:

a(5) = a(5-4) = a(1) = 2,

a(6) = a(6-4) = a(2) = -3,

a(7) = a(7-4) = a(3) = 5,

and a(8) = a(8-4) = 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 \(a_n = a_{n-4}\) 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

"\(a_n = a_{n-4}\) 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 = n-4, 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 An-4 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 A9-4 (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.