The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -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

IMO B observation:= after every four itteration the values repeat.. the total of a1+a2+a3+a4 = 3 97/4=24 and remainder 1 thus 24x3 + a1 ie 72+2=74

Can you please explain this further? How do you know that the values repeat? I tried finding the difference, but there's no consistency or pattern..

Thouraya wrote:Can you please explain this further? How do you know that the values repeat? I tried finding the difference, but there's no consistency or pattern.. an = a(n-4) => a5 = a(5-4) = a1 a9=a(9-4)=a5=a1 similarly; a1=a5=a9=a13=....a93=a97 = 2 a2=a6=a10=.....=a94 = -3 a3=a7=a11=... =a95 = 5 a4=a8=a12=.......=a96 = -1 sum = 24x(2-3+5-1) + 2 = 74

Solution: The sequence is 2, -3, 5, -1, 2, -3, 5, -1...... So, a1 = 2 = a5 = a9 = a(4k+1), where k is any integer >= 0. Similarly a2 = -3 = a6 = a10 = a(4k+2) a3 = 5 = a7 = a11 = a(4k+3) a4 = -1 = a8 = a12 = a(4k+4) Note that the sum of first 4 terms is 3 = sum of next 4 terms and so on. So, upto 96, the sum of all terms is (96/4)*3 = 72. Since 97 = 4*24+1, the 97th term is 2. Hence, the sum of 97 terms is 72+2 = 74.

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