1966 AMC 12 Problem 19

Attempt Problem 19 of the 1966 AMC 12 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 1966 AMC 12 solutions, or check the answer key.

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

19.

Let s1s_1 be the sum of the first nn terms of the arithmetic sequence 8,8, 12,12, …\ldots and let s2s_2 be the sum of the first nn terms of the arithmetic sequence 17,17, 19,19, ….\ldots. Then s1=s2s_1=s_2 for:
no value of nn
one value of nn
two values of nn
four values of nn
more than four values of nn
Answer: B
Concepts:arithmetic sequencequadraticsummation
Small hint:
Write a sum formula for each progression
Big hint:
Cancel the positive factor nn after equating the sums
Solution:
The two sums are s1=2n2+6n,s2=n2+16n. \begin{gathered} s_1=2n^2+6n,\\ s_2=n^2+16n. \end{gathered} Equality gives n(n−10)=0.n(n-10)=0. Since a number of terms is positive, only n=10n=10 works, so there is one value.
Therefore, the correct answer is B.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.