Difference between revisions of "1985 AJHSME Problems/Problem 2"
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Applying the formula, we have: | Applying the formula, we have: | ||
| − | <cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945</cmath>. | + | <cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath>. |
==Video Solution== | ==Video Solution== | ||
Revision as of 20:50, 1 February 2023
Problem
Solution 1
To simplify the problem, we can group 90’s together: 
.
, and finding 
has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
![\[(1 + 9)+(2+8)+(3+7)+(4+6)+5\]](http://latex.artofproblemsolving.com/1/5/1/151425ffc3700df99ccddfbfcf3c4452458051d3.png)
. 
, and 
.
Solution 2
We can express each of the terms as a difference from 
and then add the negatives using 
to get the answer.
Solution 3
Instead of breaking the sum then rearranging, we can rearrange directly:
Solution 4
The finite arithmetic sequence formula states that the sum in the sequence is equal to 
n
a_1
a_n$ is the last term.
Applying the formula, we have:
![\[\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}\]](http://latex.artofproblemsolving.com/9/e/7/9e7b8229c2e1de43dd8882831153b5f103b1ecf3.png)
.
Video Solution
~savannahsolver
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
