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Difference between revisions of "1975 AHSME Problems/Problem 9"

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==Problem==
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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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==Solution==
 
==Solution==
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Solution by e_power_pi_times_i
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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==
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{{AHSME box|year=1975|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 15:54, 19 January 2021

Problem

Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be arithmetic progressions such that $a_1 = 25, b_1 = 75$, and $a_{100} + b_{100} = 100$. Find the sum of the first hundred terms of the progression $a_1 + b_1, a_2 + b_2, \ldots$

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 10,000 \qquad \textbf{(D)}\ 505,000 \qquad \\ \textbf{(E)}\ \text{not enough information given to solve the problem}$


Solution

Notice that $a_{100}$ and $b_{100}$ are $25+99k_1$ and $75+99k_2$, respectively. Therefore $k_2 = -k_1$. Now notice that $a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100$. The sum of the first $100$ terms is $100\cdot100 = \boxed{\textbf{(C) } 10,000}$.

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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