Consider the sequence of numbers a1, a2, a3, ... to infinity where a1 = 81.33 and a2 = –19 and aj = aj–1 – aj–2 for j ≥ 3. What is the sum of the first 6002 terms of this sequence?
Explanation:
a1 = 81.33 a2 = –19 a3 = a2 - a1 = –100.33 a4 = a3 - a2 = –81.33 a5 = a4 - a3 = 19 a6 = a5 - a4 = 100.33 a7 = a2 - a5 = 81.33 a8 = a7 - a6 = –19
We can see that the sequence repeats itself after every 6 terms.
Sum of the first 6 terms of the sequence = 0
Thus, the sum of the first 6000 terms of this sequence = 0
The sum of the 6001st and 6002nd terms = 81.33 – 19 = 62.33
Hence, option (c).
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