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sequence

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by anishprabhu » Sat Mar 14, 2009 10:00 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

My votes on B
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Re: sequence

by DanaJ » Sat Mar 14, 2009 10:09 am
anishprabhu wrote:an = an-4 for n > 4
I'm not sure you wrote that down correctly: it doesn't make much sense... Please check.
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by anishprabhu » Sat Mar 14, 2009 10:12 am
asuffix n = a suffix(n-4) for n > 4
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by truplayer256 » Sat Mar 14, 2009 10:15 am
All the answer choices are incorrect.

a5=1
a6=2
a7=3
We want 93 of these terms since we're already given 4 terms before. Since, this is an arithmetic sequence, we can calculate the sum of 1-93 by using sn=n/2(a1+an):

93/2(1+93)=4371<---- now we add this number to 2,-3,5,-1 and we get 4374..
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by anishprabhu » Sat Mar 14, 2009 10:25 am
My solution:
a1 = 2,
a2 = -3,
a3 = 5,
a4 = -1
a5 = 2,
a6 = -3
a7 = 5,
a8 = -1 and so on.
Sequence repeats every 4 terms.

97/4 = 24 with one term remaining which would be the forst term in the sequence. thus a97 = 2.

a1+a2+a3+a4 = 3

thus 24 * 3 = 72
thus first 96 terms sum = 72
Adding the 97th term value = 72 + 2

= 74

B
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by DanaJ » Sat Mar 14, 2009 10:25 am
That basically means that a1 = a5, a2 = a6, a3 = a7, a4 = a8.
This means that if you figure out the first sequence, then the rest will be a breeze.
Notice that a1 + a2 + a3 + a4 = 2 - 3 + 5 - 1 = 3. For every sequence of 4 items, we will get the same result i.e.:
a1 + a2 + a3 + a4 = 3
a5 + a6 + a7 + a8 = 3
a9 + a10 + a11 + a12 = 3
....
a93 + a94 + a95 + a96 = 3
You have 96/4 = 24 sequences of 3, so you get 24*3 = 72 for the first 96 items. a97 will be a1 = 2, so the total will be 72 + 2 = 74.
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by maihuna » Sat Mar 14, 2009 11:46 am
what is the source of this question?
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