Difference between revisions of "1980 AHSME Problems/Problem 11"
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== Solution == | == Solution == | ||
| − | <math>\ | + | Let <math>a</math> be the first term of the sequence and let <math>d</math> be the common difference of the sequence. |
| + | |||
| + | Sum of the first 10 terms: <math>\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20</math> | ||
| + | Sum of the first 100 terms: <math>\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}</math> | ||
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| + | Solving the system, we get <math>d=-\frac{11}{50}</math>, <math>a=\frac{1099}{100}</math>. The sum of the first 110 terms is <math>\frac{110}{2}(2a+109d)=55(-2)=-110</math> | ||
| + | |||
| + | Therefore, <math>\boxed{D}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 14:31, 22 April 2016
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 
be the first term of the sequence and let 
be the common difference of the sequence.
Sum of the first 10 terms: 
Sum of the first 100 terms: 
Solving the system, we get 
, 
. The sum of the first 110 terms is 
Therefore, 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
