1969 AHSME Problems/Problem 33
Problem
Let and be the respective sums of the first terms of two arithmetic series. If for all , the ratio of the eleventh term of the first series to the eleventh term of the second series is:
Solution
Let be the first arithmetic sequence and be the second arithmetic sequence. If , then . Since and are just the first term, the first term of is and the first term of is for some . If , then , so the sum of the first two terms of is and the sum of the first two terms of is for some . Thus, the second term of is and the second term of is , so the common difference of is and the common difference of is .
Thus, using the first terms and common differences, the sum of the first three terms of equals , and the sum of the first three terms of equals . That means With the substitution, the common difference of is , and the common difference of is . That means the term of is , and the term of is . Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is .
Solution 2 (Quick Sol)
Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of . From the ratio given in the problem, multiply to both sides to get quadratic polynomials. From there, the 11th term for and can becalculated. The ratio is .
Solution 3 (Quickest Sol)
Since they are arithmetic sequences, we know that the 11th term is the average value of the first 21 terms in each sequence. So the desired ratio of the 11th terms is just
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See Also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
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