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trouble with sequence

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by willbeatthegmat » Wed Jan 21, 2009 8:01 am
Q- 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 amitabhprasad » Wed Jan 21, 2009 8:35 am
I am getting answer as 74 ?
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by naaga » Wed Jan 21, 2009 8:41 am
amitabh can you please explain in detail .....
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by amitabhprasad » Wed Jan 21, 2009 8:52 am
naaga wrote:amitabh can you please explain in detail .....
Sure,
a1 = 2, a2 = -3, a3 = 5, a4 = -1
==> a1+a2+a3+a4 = 3
now we have an = an-4
==> a5 = a5-4 = a1=2
a6 = a6-4 =a2 = -3
a7 = a7-4 = a3 = 5
a8 = a8-4 = a4 = -1
and so on you will see we get sequence in the block of 4
We are supposed to find the sum of first 97 terms
so sum of 97 terms = 24*(sum of 1st 4 terms )+1st term
=24*3+2 = 74
HTH
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by gaggleofgirls » Thu Jan 22, 2009 10:29 am
For the infinite sequence, the terms 2, -3, 5, -1 repeat infinitely.

2 + -3 + 5 -1 = 3, so the sum for every set of the 4 numbers is 4.

How many sets of these 4 numbers are there in the first 97 terms?

97/4 = 24 remainder 1

So, first let's address the 24. There are 24 sets of numbers whose sum is 3 in the first 97 terms, 24 * 3 = 72.

But now you have to address the remainder 1, there is 1 more number in the sequence, and it will be the next number in line, which is the first number of the set of 4, which is a1, which is 2.

So, the answer is 72 +2 = 74

Answer is B

-Carrie
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