1975 AHSME Problems/Problem 15

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Problem

In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is

$\textbf{(A)}\ 5 \qquad  \textbf{(B)}\ 4 \qquad  \textbf{(C)}\ 2 \qquad  \textbf{(D)}\ 1 \qquad  \textbf{(E)}\ -1$

Solution

First, write a few terms of the sequence: $1, 3, 2, -1, -3, -2, 1, 3, 2, \ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$th term is $-2$ and the sum of the all those previous terms is $0$. Then, write the 97th to the 100th terms down: $1, 3, 2,-1$ and add them up to get the sum of $\boxed{\textbf{(A) } 5}$

See Also

2019 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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