1984 AIME Problems/Problem 1

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

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1984 AIME Problems/Problem 1

Problem

Solution

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