Difference between revisions of "1961 AHSME Problems/Problem 26"
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Let the first term of the [[arithmetic sequence]] be <math>a</math> and the common difference be <math>d</math>. | Let the first term of the [[arithmetic sequence]] be <math>a</math> and the common difference be <math>d</math>. | ||
− | The <math>50^{\text{th}}</math> term of the sequence is <math>a+49d</math>, so the sum of the first <math>50</math> terms is <math>\frac{50(a + a + 49d}{2}</math>. | + | The <math>50^{\text{th}}</math> term of the sequence is <math>a+49d</math>, so the sum of the first <math>50</math> terms is <math>\frac{50(a + a + 49d)}{2}</math>. |
The <math>51^{\text{th}}</math> term of the sequence is <math>a+50d</math> and the <math>100^{\text{th}}</math> term of the sequence is <math>a+99d</math>, so the sum of the next <math>50</math> terms is <math>\frac{50(a+50d+a+99d)}{2}</math>. | The <math>51^{\text{th}}</math> term of the sequence is <math>a+50d</math> and the <math>100^{\text{th}}</math> term of the sequence is <math>a+99d</math>, so the sum of the next <math>50</math> terms is <math>\frac{50(a+50d+a+99d)}{2}</math>. |
Latest revision as of 10:38, 22 May 2018
Problem
For a given arithmetic series the sum of the first terms is , and the sum of the next terms is . The first term in the series is:
Solution
Let the first term of the arithmetic sequence be and the common difference be .
The term of the sequence is , so the sum of the first terms is .
The term of the sequence is and the term of the sequence is , so the sum of the next terms is .
Substituting in values results in this system of equations. Solving for yields . The first term is , which is answer choice .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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All AHSME Problems and Solutions |
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