Difference between revisions of "1974 AHSME Problems/Problem 29"
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{{AHSME box|year=1974|num-b=28|num-a=30}} | {{AHSME box|year=1974|num-b=28|num-a=30}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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Latest revision as of 12:44, 5 July 2013
Problem
For 
let 
be the sum of the first 
terms of the arithmetic progression whose first term is 
and whose common difference is 
; then 
is
Solution
The 
term of an arithmetic progression with a first term and a common difference is 
. Therefore, the sum of the first terms of such a progression is 
.
We now want to evaluate 
. ![\[\sum_{p=1}^{10}(1600p-780)=1600\sum_{p=1}^{10}(p)-\sum_{p=1}^{10}(780)\]](http://latex.artofproblemsolving.com/9/f/6/9f62acd412c822c6caf5f9f07ca281051bd9badd.png)
![\[=(1600)\left(\frac{10\cdot11}{2}\right)-(780)(10)=88000-7800=80200, \boxed{\text{B}}.\]](http://latex.artofproblemsolving.com/9/1/9/919b65da8dfd8dea96fa0c786e11d6f36ff74ec4.png)
See Also
| 1974 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 28 |
Followed by Problem 30 | |
| 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. 
