1985 AJHSME Problems/Problem 2
Problem
Solution 1
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. We find a simpler problem in this problem, and simplify ->
We know , that's easy: . So how do we find ?
We rearrange the numbers to make . You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. . Adding that on to 900 makes 945.
945 is
Solution 2
We can express each of the terms as a difference from 100 and then add the negatives using to get the answer.
Solution 3
Instead of breaking the sum and then rearranging, we can start by rearranging:
Solution 4
We can use the formula for finite arithmetic sequences.
It is () where is the number of terms in the sequence, is the first term and is the last term.
Applying it here:
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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