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

(Created page with "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</ma...")
 
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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.
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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>.

Revision as of 12:18, 16 December 2016

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 $100\cdot100 = \boxed{\textbf{(C) } 10,000}$.

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