3, 5,–5...
The first term in the sequence of numbers shown above is 3. Each even–numbered term is 2 more than the previous term and each odd–numbered term, after the first, is –1 times the previous term. For example, the second term is 3 + 2, and the third term is (–1) × 5. What is the 55th term of the sequence?
(A)–5 (B) –3 (C) –1 (D) 3 (E) 5
1 is odd, and 1 more than a multiple of 4, since 1=4*0+1, and the 1st term is 3
2 is even, but an even NON-multiple of 4, and the 2nd term is 3+2 or 5
3 is odd, and 1 more than an even NON-multiple of 4 and the 3rd term is -5
4 is even, a multiple of 4, since 4=4*1, and the 4th term is -5+2 or -3
5 is odd, and 1 more than a multiple of 4, and the 5th term is 3
And we're back to where we started, so the rule for the sequence is
I. If n is odd, and 1 more than a multiple of 4, then the nth term is 3
II. If n is odd, and 1 more than an even NON-multiple of 4, then the nth term is -5
III. If n is even, but a NON-multiple of 4, the nth term is 5
IV. If n is even, and a multiple of 4, the nth term is -3
We want the 55th term. 55 is odd, and is 1 more than 54, an even but
NON-multiple of 4, so the 55th term is in category II above, and is
therefore -5
The correct choice is (A) -5
Edwin