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

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:

$\frac{137 - 97(49) + 2(49)^2}{2} = \fbox{093}$.

Solution 3

A better approach to this problem is to notice that from $a_{1}+a_{2}+\cdots a_{98}=137$ that each element with an odd subscript is 1 from each element with an even subscript. Thus, we note that the sum of the odd elements must be $\frac{137-49}{2}$. Thus, if we want to find the sum of all of the even elements we simply add $49$ common differences to this giving us $\frac{137-49}{2}+49=\fbox{93}$.

See also

1984 AIME (ProblemsAnswer KeyResources)
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
ACS WASC
ACCREDITED
SCHOOL

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