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1988 AJHSME Problems/Problem 19

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Problem

What is the 100\text{th} number in the arithmetic sequence: 1,5,9,13,17,21,25,...?

\text{(A)}\ 397 \qquad \text{(B)}\ 399 \qquad \text{(C)}\ 401 \qquad \text{(D)}\ 403 \qquad \text{(E)}\ 405

Solution

To get from the 1^\text{st} term of an arithmetic sequence to the 100^\text{th} term, we must add the common difference 99 times. The first term is 1 and the common difference is 5-1=9-5=13-9=\cdots = 4, so the 100^\text{th} term is 1+4(99)=397 \rightarrow \boxed{\text{A}}

See Also

1988 AJHSME (ProblemsResources)
Preceded by
Problem 18
Followed by
Problem 20
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
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