Difference between revisions of "1984 AIME Problems/Problem 1"
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One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer. | One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to find the value of <math>a_1</math>, then use that to calculate <math>a_2</math> and sum another arithmetic series to get our answer. | ||
− | A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>. The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = 093</math>. | + | A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substitute this into our given equation to get <math>(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137</math>. The left-hand side of this equation is simply <math>2(a_2 + a_4 + \ldots + a_{98}) - 49</math>, so our desired value is <math>\frac{137 + 49}{2} = \boxed{093}</math>. |
== See also == | == See also == |
Revision as of 17:04, 21 March 2009
Problem
Find the value of if , , is an arithmetic progression with common difference 1, and .
Solution
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 .
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 |