Difference between revisions of "1969 AHSME Problems/Problem 33"
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== Solution == | == Solution == | ||
− | <math> | + | Let <math>S</math> be the first arithmetic sequence and <math>T</math> be the second arithmetic sequence. If <math>n = 1</math>, then <math>S_1:T_1 = 8:31</math>. Since <math>S_1</math> and <math>T_1</math> are just the first term, the first term of <math>S</math> is <math>8a</math> and the first term of <math>T</math> is <math>31a</math> for some <math>a</math>. If <math>n = 2</math>, then <math>S_2:T_2 = 15:35 = 3:7</math>, so the sum of the first two terms of <math>S</math> is <math>3b</math> and the sum of the first two terms of <math>T</math> is <math>7b</math> for some <math>b</math>. Thus, the second term of <math>S</math> is <math>3b-8a</math> and the second term of <math>T</math> is <math>7b - 31a</math>, so the common difference of <math>S</math> is <math>3b-16a</math> and the common difference of <math>T</math> is <math>7b-62a</math>. |
− | == See | + | Thus, using the first terms and common differences, the sum of the first three terms of <math>S</math> equals <math>\tfrac{1}{2} \cdot 3(16a + 2(-16a + 3b))</math>, and the sum of the first three terms of <math>T</math> equals <math>\tfrac{1}{2} \cdot 3(62a + 2(-62a + 7b))</math>. That means |
+ | <cmath>\frac{16a + 2(-16a + 3b)}{62a + 2(-62a + 7b)} = \frac{22}{39}</cmath> | ||
+ | <cmath>\frac{6b-16a}{14b-62a} = \frac{22}{39}</cmath> | ||
+ | <cmath>\frac{3b-8a}{7b-31a} = \frac{22}{39}</cmath> | ||
+ | <cmath>117b - 312a = 154b - 682a</cmath> | ||
+ | <cmath>-37b = -370a</cmath> | ||
+ | <cmath>b = 10a</cmath> | ||
+ | With the substitution, the common difference of <math>S</math> is <math>14a</math>, and the common difference of <math>T</math> is <math>8a</math>. That means the <math>11^\text{th}</math> term of <math>S</math> is <math>8a + 10(14a) = 148a</math>, and the <math>11^\text{th}</math> term of <math>T</math> is <math>31a + 10(8a) = 111a</math>. Thus, the ratio of the eleventh term of the first series to the eleventh term of the second series is <math>148a:111a = \boxed{\textbf{(A) } 4:3}</math>. | ||
+ | |||
+ | == Solution 2 (Quick Sol) == | ||
+ | Note that the sum of arithmetic sequences can be expressed as a quadratic polynomial of <math>n</math>. From the ratio given in the problem, multiply <math>n</math> to both sides to get quadratic polynomials. <cmath>S_n = 7n^2+n, T_n = 4n^2+27n</cmath> From there, the 11th term for <math>S_n</math> and <math>T_n</math> can becalculated. <cmath>S_{11} - S_{10} = 7*11^2+11 - (7*10^2+10) = 148</cmath> <cmath>T_{11} - T_{10} = 4*11^2+27*11 - (4*10^2+27*10) = 111</cmath> The ratio is <math>148 : 111 = \boxed{\textbf{(A) } 4:3}</math>. | ||
+ | |||
+ | == See Also == | ||
{{AHSME 35p box|year=1969|num-b=32|num-a=34}} | {{AHSME 35p box|year=1969|num-b=32|num-a=34}} | ||
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:51, 1 June 2021
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 .
See Also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
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