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

Revision as of 21:25, 13 May 2011

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

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 $a_1$ , 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 $\frac{137 + 49}{2} = \boxed{093}$ .

Solution 2

If $a_1$ is the first term, then $a_1+a_2+a_3 + \cdots + a_{98} = 137$ can be rewritten as:

$98a_1 + 1+2+3+ \cdots + 97 = 137$ $\Leftrightarrow$ $98a_1 + \frac{97 \cdot 98}{2} = 137$

Our desired value is $a_2+a_4+a_6+ \cdots + a_{98}$ so this is:

$49a_1 + 1+3+5+ \cdots + 97$

which is $49a_1+ 49^2$ . So, from the first equation, we know $49a_1 = \frac{137}{2} - \frac{97 \cdot 49}{2}$ . So, the final answer is:

$\frac{137 - 97(49) + 2(49)^2}{2} = \fbox{93}$ .

@@ Line 14: / Line 14: @@
 If <math> a_1 </math> is the first term, then <math> a_1+a_2+a_3 + \cdots + a_{98} = 137 </math> can be rewritten as:
-<math> 98a_1 + 1+2+3+ \cdots + 97 = 137 </math>
+<math> 98a_1 + 1+2+3+ \cdots + 97 = 137 </math> <math>\Leftrightarrow</math>
 <math> 98a_1 + \frac{97 \cdot 98}{2} = 137 </math>

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
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