Difference between revisions of "1975 AHSME Problems/Problem 9"
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Let <math>a_1, a_2, \ldots</math> and <math>b_1, b_2, \ldots</math> be arithmetic progressions such that <math>a_1 = 25, b_1 = 75</math>, and <math>a_{100} + b_{100} = 100</math>. | Let <math>a_1, a_2, \ldots</math> and <math>b_1, b_2, \ldots</math> be arithmetic progressions such that <math>a_1 = 25, b_1 = 75</math>, and <math>a_{100} + b_{100} = 100</math>. | ||
Find the sum of the first hundred terms of the progression <math>a_1 + b_1, a_2 + b_2, \ldots</math> | Find the sum of the first hundred terms of the progression <math>a_1 + b_1, a_2 + b_2, \ldots</math> | ||
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Notice that <math>a_{100}</math> and <math>b_{100}</math> are <math>25+99k_1</math> and <math>75+99k_2</math>, respectively. Therefore <math>k_2 = -k_1</math>. Now notice that <math>a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100</math>. The sum of the first <math>100</math> terms is <math>100\cdot100 = \boxed{\textbf{(C) } 10,000}</math>. | Notice that <math>a_{100}</math> and <math>b_{100}</math> are <math>25+99k_1</math> and <math>75+99k_2</math>, respectively. Therefore <math>k_2 = -k_1</math>. Now notice that <math>a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100</math>. The sum of the first <math>100</math> terms is <math>100\cdot100 = \boxed{\textbf{(C) } 10,000}</math>. | ||
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+ | ==See Also== | ||
+ | {{AHSME box|year=1975|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:54, 19 January 2021
Problem
Let and be arithmetic progressions such that , and . Find the sum of the first hundred terms of the progression
Solution
Notice that and are and , respectively. Therefore . Now notice that . The sum of the first terms is .
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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All AHSME Problems and Solutions |
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