Difference between revisions of "1984 AIME Problems/Problem 1"
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which is <math> 49a_1+ 49^2 </math>. So, from the first equation, we know <math> 49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2} </math>. So, the final answer is: | which is <math> 49a_1+ 49^2 </math>. So, from the first equation, we know <math> 49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2} </math>. So, the final answer is: | ||
− | <math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{ | + | <math> \frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093} </math>. |
== See also == | == See also == |
Revision as of 21:26, 13 May 2011
Problem
Find the value of if , , is an arithmetic progression with common difference 1, and .
Solution
Solution 1
One approach to this problem is to apply the formula for the sum of an arithmetic series in order to find the value of , then use that to calculate and sum another arithmetic series to get our answer.
A somewhat quicker method is to do the following: for each , we have . We can substitute this into our given equation to get . The left-hand side of this equation is simply , so our desired value is .
Solution 2
If is the first term, then can be rewritten as:
Our desired value is so this is:
which is . So, from the first equation, we know . So, the final answer is:
.
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |