1984 AIME Problems/Problem 1

Problem

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

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 $\frac{137 + 49}{2} = \boxed{93}$ .

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:

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

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 $\frac{137-49}{2}+49=\fbox{93}$ .

Or, since the sum of the odd elements is 44, then the sum of the even terms must be $\fbox{93}$ .

Solution 4

We want to find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ , which can be rewritten as $a_1+1+a_2+2+a_3+\ldots+a_{49}+49 \implies a_1+a_2+a_3+\ldots+a_{49}+\frac{49 \cdot 50}{2}$ . We can split $a_1+a_2+a_3+\ldots+a_{98}$ into two parts: $a_1+a_2+a_3+\ldots+a_{49}$ and $a_{50}+a_{51}+a_{52}+\ldots+a_{98}$ Note that each term in the second expression is $49$ greater than the corresponding term, so, letting the first equation be equal to $x$ , we get $a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}$ . Calculating $49^2$ by sheer multiplication is not difficult, but you can also do $(50-1)(50-1)=2500-100+1=2401$ . We want to find the value of $x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225$ . Since $x=\frac{137-2401}{2}$ , we find $x=-1132$ . $-1132+1225=\boxed{93}$ .

- PhunsukhWangdu

Solution 5

Since we are dealing with an arithmetic sequence, $a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}$ We can also figure out that $a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137$ $a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137$ Thus, $49a_{50} = \frac{137 + 49}{2} = \boxed{93}$

~kempwood

Video Solution by OmegaLearn

https://youtu.be/re8DTg-Lbu0?t=234

~ pi_is_3.14

1984 AIME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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