1988 AMC 8 Problem 19 — Solution | LIVE by Po-Shen Loh

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1988 AMC 8 Problem 19

Below is the professionally curated solution for Problem 19 of the 1988 AMC 8, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:arithmetic sequence

Difficulty rating: 820

19.

What is the $100$ th number in the arithmetic sequence $1, 5, 9, 13, 17, 21, 25, \ldots?$

$397$

$399$

$401$

$403$

$405$

Solution:

The sequence starts at $1$ with common difference $4.$ The $100$ th term is reached by adding $4$ to the first term $99$ times.

So the $100$ th term is $1 + 99 \times 4 = 1 + 396 = 397.$

Thus, the correct answer is A .

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