1952 AMC 12 Problem 30

Attempt Problem 30 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

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30.

When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is:
1:21:2
2:12:1
1:41:4
4:14:1
1:11:1
Answer: A
Concepts:arithmetic sequencesummationratio and proportion
Small hint:
Write both partial sums in terms of the first term aa and common difference dd
Big hint:
Use Sn=n2(2a+(n−1)d)S_n=\dfrac n2(2a+(n-1)d) and simplify S10=4S5S_{10}=4S_5
Solution:
The sum formula gives S10=5(2a+9d),S5=52(2a+4d). \begin{aligned} S_{10}&=5(2a+9d),\\ S_5&=\frac52(2a+4d). \end{aligned} The equation S10=4S5S_{10}=4S_5 becomes 2a+9d=2(2a+4d), 2a+9d=2(2a+4d), so d=2a.d=2a. Hence a:d=1:2.a:d=1:2.
Thus, the correct answer is A.
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