1955 AMC 12 Problem 45

Attempt Problem 45 of the 1955 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 1955 AMC 12 solutions, or check the answer key.

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45.

Given a geometric sequence with the first term 0\ne0 and r0r\ne0 and an arithmetic sequence with the first term =0.=0. A third sequence 1,1, 1,1, 2,2, \ldots is formed by adding corresponding terms of the two given sequences. The sum of the first ten terms of the third sequence is:

978978

557557

467467

10681068

not possible to determine from the information given

Answer: A
Concepts:geometric sequencearithmetic sequencesystems of equations
Difficulty rating: 1930
Small Hint:

Write the two sequences as a,ar,ar2,a,ar,ar^2,\ldots and 0,d,2d,0,d,2d,\ldots

Big Hint:

Use the first three sums and the condition r0r\ne0 to determine a,r,da,r,d

Solution:

The first three termwise sums give a=1,r+d=1,r2+2d=2. \begin{aligned} a&=1,\\ r+d&=1,\\ r^2+2d&=2. \end{aligned} Substituting d=1rd=1-r yields r(r2)=0.r(r-2)=0. Since r0,r\ne0, we have r=2r=2 and d=1.d=-1. The first ten geometric terms sum to 2101=1023,2^{10}-1=1023, while the first ten arithmetic terms sum to 019=45.0-1-\cdots-9=-45. Their combined sum is 102345=978.1023-45=978.

Thus, the correct answer is A.

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Problem 45 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12