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Difficulty: 505-555 Level,    Sequences,                            
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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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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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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Sum of a sequence [#permalink]
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\(a_1 = 2\)

\(a_2 = -3\)

\(a_3 = 5\)

\(a_4 = -1\)

\(a_5 = a_{(5-4)} = a_1 = 2\)

\(a_6 = a_{(6-4)} = a_2 = -3\)

\((a_1 + a_2 + a_3 + a_4) = (a_5 + a_6 + a_7 + a_8) = (a_9 + a_{10} .....)\) 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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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


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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The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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


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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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 \(a_n = a_{n-4}\) 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


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]
Bunuel wrote:
udaymathapati wrote:
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 \(24*3+2=74\).

Answer: B.



Hi can you help understand why can't we use the formula here N(A1+AN)/2 ? Thanks
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Re: The infinite sequence a1, a2, …, an, … is such that a1=2 [#permalink]
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Sri07 wrote:
Bunuel wrote:
udaymathapati wrote:
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 \(24*3+2=74\).

Answer: B.



Hi can you help understand why can't we use the formula here N(A1+AN)/2 ? Thanks


The formula you quote, is for evenly spaced sets (aka arithmetical progression), and cannot be applied to all sequences. The sequence at hand is 2, -3, , 5, -1, ... As you can see this is not an arithmetic progression, so you cannot use that formula.

Check links below for more.


12. Sequences



For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: The infinite sequence a1, a2, , an, is such that a1=2 [#permalink]
Hello,

So this progression is neither an AP or GP correct? and thus the equations for the sum do not apply
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Re: The infinite sequence a1, a2, , an, is such that a1=2 [#permalink]
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