Question

Consider the sequence of numbers \(a_1, a_2, a_3, \dots \) to infinity where \(a_1 = 81.33\) and \(a_2 = -19\) and \(a_j = a_{j-1} - a_{j-2}\) for \(j \geq 3\). What is the sum of the first 6002 terms of this sequence?

• A. -100.33
• B. -30.00
• C. 62.33
• D. 119.33

Hint:

In recurrence sequence problems, first look for periodicity or patterns that simplify the sum over many terms.

Answer 1

The Correct Option is C

Given the recurrence relation \(a_j = a_{j-1} - a_{j-2}\), we see that the terms of the sequence will alternate. By calculating the terms and considering the periodicity of the sequence, the sum of the first 6002 terms is 62.33. \[ \boxed{62.33} \]