Difference between revisions of "2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"
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== Solution== | == Solution== | ||
Let <math>d_1</math> be the common difference in the first sequence and <math>d_2</math> the common difference in the second sequence. Thus, <math>b_4=x+4d_2</math> and <math>b_3=x+2d_2</math>. In addition, <math>a_2=x+2d_1</math> and <math>a_1=x+d_1</math>. Consequently <cmath>\frac{b_4-b_3}{a_2-a_1}=\frac{x+4d_2-x-2d_2}{x+2d_1-x-d_1}</cmath> or <cmath>\frac{b_4-b_3}{a_2-a_1}=\frac{2d_1}{d_2}</cmath> Since <math>d_2=\frac{y-x}{3}</math> and <math>d_1=\frac{y-x}{4}</math>, we have <cmath>\frac{2d_1}{d_2}=\frac{\frac{2y-2x}{3}}{\frac{y-x}{4}}</cmath> which simplifies to <math>\boxed{\frac83}</math>. | Let <math>d_1</math> be the common difference in the first sequence and <math>d_2</math> the common difference in the second sequence. Thus, <math>b_4=x+4d_2</math> and <math>b_3=x+2d_2</math>. In addition, <math>a_2=x+2d_1</math> and <math>a_1=x+d_1</math>. Consequently <cmath>\frac{b_4-b_3}{a_2-a_1}=\frac{x+4d_2-x-2d_2}{x+2d_1-x-d_1}</cmath> or <cmath>\frac{b_4-b_3}{a_2-a_1}=\frac{2d_1}{d_2}</cmath> Since <math>d_2=\frac{y-x}{3}</math> and <math>d_1=\frac{y-x}{4}</math>, we have <cmath>\frac{2d_1}{d_2}=\frac{\frac{2y-2x}{3}}{\frac{y-x}{4}}</cmath> which simplifies to <math>\boxed{\frac83}</math>. | ||
+ | - Juno | ||
== See also == | == See also == |
Latest revision as of 13:56, 6 May 2021
Problem
Let be two real numbers. Let and be two arithmetic sequences.
Calculate .
Solution
Let be the common difference in the first sequence and the common difference in the second sequence. Thus, and . In addition, and . Consequently or Since and , we have which simplifies to . - Juno
See also
2018 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |