Difference between revisions of "1980 AHSME Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | Let <math>a</math> | + | Let <math>a</math> and <math>d</math> be the first term and common difference of the sequence, respectively. |
The sum of the first <math>10</math> terms is <math>\frac{10}{2}(2a+9d)=100</math>. This equation can be simplified to <math>2a+9d=20</math>. | The sum of the first <math>10</math> terms is <math>\frac{10}{2}(2a+9d)=100</math>. This equation can be simplified to <math>2a+9d=20</math>. | ||
Revision as of 20:54, 25 July 2025
Problem
If the sum of the first 
terms and the sum of the first 
terms of a given arithmetic progression are and ,
respectively, then the sum of first 
terms is:
Solution
Let 
and 
be the first term and common difference of the sequence, respectively.
The sum of the first terms is 
. This equation can be simplified to 
.
The sum of the first terms is 
. This equation can be simplified to 
.
Solving the system of these two equations yields 
. Therefore, the sum of the first terms is 
.
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
