1961 AMC 12 Problem 26

Attempt Problem 26 of the 1961 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 1961 AMC 12 solutions, or check the answer key.

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

26.

For a given arithmetic series the sum of the first 5050 terms is 200,200, and the sum of the next 5050 terms is 2700.2700. The first term in the series is:
−1212-12\dfrac12
−21.5-21.5
−20.5-20.5
33
3.53.5
Answer: C
Concepts:arithmetic sequencesummationsystem of equations
Small hint:
Write the first sum in terms of the first term aa and common difference dd
Big hint:
The next 5050 terms have first term a+50da+50d
Solution:
The two sums give 25(2a+49d)=200,25(2a+149d)=2700. \begin{aligned} 25(2a+49d)&=200,\\ 25(2a+149d)&=2700. \end{aligned} Thus 2a+49d=82a+49d=8 and 2a+149d=108.2a+149d=108. Subtraction gives d=1,d=1, and then 2a=−41,2a=-41, so a=−20.5.a=-20.5.
Therefore, the correct answer is C.
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