Difference between revisions of "1985 AJHSME Problems/Problem 2"
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==Solution 4== | ==Solution 4== | ||
| − | + | The finite arithmetic sequence formula states that the sum in the sequence is equal to <math>\frac{n}{2}\cdot(a_1+a_n)3 where </math>n<math> is the number of terms in the sequence, </math>a_1<math> is the first term and </math>a_n$ is the last term. | |
| − | + | Applying the formula, we have: | |
| − | + | <cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945</cmath>. | |
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==Video Solution== | ==Video Solution== | ||
Revision as of 20:49, 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:
\[\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945\] (Error compiling LaTeX. Unknown error_msg).
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
