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by Newaz111 » Wed May 27, 2015 11:52 am
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
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by Spencer@Prep4GMAT » Wed May 27, 2015 12:07 pm
Newaz111 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
Since a(n) = a(n-4), the next term a(5) = a(1) = 2. We have a repeating sequence 2, -3, 5, -1, 2, -3, 5, -1... The sum of every 4 terms is 2 - 3 + 5 - 1 = 3. The first 96 terms contain 96/4 = 24 sets of these 4 terms, so the sum of the first 96 terms is 24(3) = 72. The 97th term begins the sequence again with 2, so adding the 97th term gives us 72 + 2 = 74.
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by nikhilgmat31 » Mon Jun 01, 2015 9:25 pm
24 * (2 -3 + 5 -1 ) + 2
= 24 * 3 + 2
74 indeed.
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by GMATinsight » Tue Jun 02, 2015 2:52 am
Newaz111 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
Here, an = an-4 for n > 4

represents that 5th terms of the series will be same as 1st Term and
6th terms of the series will be same as 2nd Term and
7th terms of the series will be same as 3rd Term and
8th terms of the series will be same as 4th Term and so on...

i,e, terms will keep repeating and the series of sets will be

{2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1, ...and so on...}

The Cyclicity of terms is 4 n this case as the terms repeat after the cycle of 4 terms

Therefore for 97th terms let's divide 97/4 in order to calculate all repeating cycles
On dividing 97 by 4, we get quotient 26 and remainder 1 i.e. 1 terms remaining after all the repetition of 24 cycles

Sum of 1 cycle of 4 terms = 2-3+5-1 = 3
Sum of 24 cycle of 4 terms each = 3 x 24 = 72
i.e. 97th terms will be same as 1st term i.e. 2

Sum of all 97 terms = 72+2 = 74

Answer: Option B
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by GMATinsight » Tue Jun 02, 2015 2:52 am
Newaz111 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
Here, an = an-4 for n > 4

represents that 5th terms of the series will be same as 1st Term and
6th terms of the series will be same as 2nd Term and
7th terms of the series will be same as 3rd Term and
8th terms of the series will be same as 4th Term and so on...

i,e, terms will keep repeating and the series of sets will be

{2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1, ...and so on...}

The Cyclicity of terms is 4 n this case as the terms repeat after the cycle of 4 terms

Therefore for 97th terms let's divide 97/4 in order to calculate all repeating cycles
On dividing 97 by 4, we get quotient 26 and remainder 1 i.e. 1 terms remaining after all the repetition of 24 cycles

Sum of 1 cycle of 4 terms = 2-3+5-1 = 3
Sum of 24 cycle of 4 terms each = 3 x 24 = 72
i.e. 97th terms will be same as 1st term i.e. 2

Sum of all 97 terms = 72+2 = 74

Answer: Option B
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by scottchapman » Tue Jun 02, 2015 2:42 pm
GMATinsight:

Could you explain/break this concept down for me?

Here, an = an-4 for n > 4
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by GMATinsight » Tue Jun 02, 2015 5:55 pm
scottchapman wrote:GMATinsight:

Could you explain/break this concept down for me?

Here, an = an-4 for n > 4
Here an represents nth term
i.e. a1 represents 1st term of the series
ie. a2 represents 2nd term of the series
ie. a3 represents 3rd term of the series
ie. a4 represents 4th term of the series and so on...

Since an = an-4 for n>4
so we can start substituting values of n.4 and get next terms if the series as mentioned below
So for n=5, a5 = a5-4 = a1
So for n=6, a6 = a6-4 = a2
So for n=7, a7 = a7-4 = a3
So for n=8, a8 = a8-4 = a4
...and so on...


I hope it's clear now...
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