1988 AMC 8 Problem 19

Attempt Problem 19 of the 1988 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

What is the 100100th number in the arithmetic sequence 1,1, 5,5, 9,9, 13,13, 17,17, 21,21, 25,25, …?\ldots?
397397
399399
401401
403403
405405
Answer: A
Concepts:arithmetic sequence
Small hint:
The common difference is 4,4, so each term adds 44 to the previous one
Big hint:
Starting from 1,1, reaching the 100100th term means adding 44 a total of 9999 times
Solution:
The sequence starts at 11 with common difference 4.4. The 100100th term is reached by adding 44 to the first term 9999 times.
So the 100100th term is 1+99×4=1+396=397.1 + 99 \times 4 = 1 + 396 = 397.
Thus, the correct answer is A .
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