AIME 1984 · 第 1 题

Solution 1

One approach is to sum an arithmetic series, find the value of $a_1$, calculate $a_2$, and sum another arithmetic series to get the answer. A quicker method is to substitue $a_{2n - 1} = a_{2n} - 1$ into the given equation:

$$(a_2 - 1) + a_2 + (a_4 - 1) + a_4 + \ldots + (a_{98} - 1) + a_{98} = 137$$ Simplify the LHS:

$$2(a_2 + a_4 + \ldots + a_{98}) - 49= 137$$ Solve:

$$a_2 + a_4 + \ldots + a_{98}=\frac{137 + 49}{2} = \boxed{93}$$

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