Difference between revisions of "1984 AIME Problems/Problem 1"

Revision as of 21:23, 5 March 2007

Problem

Find the value of $\displaystyle a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $\displaystyle a_1$ , $\displaystyle a_2$ , $\displaystyle a_3\ldots$ is an arithmetic progression with common difference 1, and $\displaystyle a_1+a_2+a_3+\ldots+a_{98}=137$ .

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 $a_1$ , then use that to calculate $a_2$ and sum another arithmetic series to get our answer.

A somewhat quicker method is to do the following: for each $n \geq 1$ , we have $a_{2n - 1} = a_{2n} - 1$ . We can substitute this into our given equation to get $(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$ . The left-hand side of this equation is simply $2(a_2 + a_4 + \ldots + a_{98}) - 49$ , so our desired value is $\frac{137 + 49}{2} = 093$ .

@@ Line 8: / Line 8: @@
 == See also ==
-* [[1984 AIME Problems/Problem 2 | Next problem]]
+{{AIME box|year=1984|before=First Question|num-a=2}}
-* [[1984 AIME Problems]]
 [[Category:Intermediate Algebra Problems]]

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

Page

Toolbox

Search

Difference between revisions of "1984 AIME Problems/Problem 1"

Revision as of 21:23, 5 March 2007

Problem

Solution

See also