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
sequence problem
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If an = an-4 for all n greater than 4, then a5 = a1, a6 = a2, a7 = a3, a8 = a4, a9 = a5 = a1, etc. So the sequence is just:zagcollins 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
2, -3, 5, -1, 2, -3, 5, -1, 2, -3, 5, -1 ....
This repeats in blocks of four. The sum of the first four terms is 3. If we add the first 96 terms, we'll just be adding that same block of four terms 24 times; the sum will be 24*3 = 72. If we add on the 97th term, which will be 2, we get the final answer: 72 + 2 = 74.
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nice explanation Ian.. i actually thought it an - 4 rather its an-4( as in a base (n-4))
thanks for the answer.
thanks for the answer.
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I too thought as below. But, once I saw a part of explanation from I realized that it is meant to be base (n-4). Then it was easy to solve
sudhir3127 wrote:nice explanation Ian.. i actually thought it an - 4 rather its an-4( as in a base (n-4))
thanks for the answer.
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I too thought the same thing and was confused. The moment I realized that it was meant to be base(n-4), it was easy to solve
sudhir3127 wrote:nice explanation Ian.. i actually thought it an - 4 rather its an-4( as in a base (n-4))
thanks for the answer.