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 (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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