Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4)
Line 34: Line 34:
 
We can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts:
 
We can split <math>a_1+a_2+a_3+\ldots+a_{98}</math> into two parts:
 
<cmath>a_1+a_2+a_3+\ldots+a_{49}</cmath> and <cmath>a_{50}+a_{51}+a_{52}+\ldots+a_{98}</cmath>
 
<cmath>a_1+a_2+a_3+\ldots+a_{49}</cmath> and <cmath>a_{50}+a_{51}+a_{52}+\ldots+a_{98}</cmath>
−
Note that each term in the second expression is <math>49</math> greater than the corresponding term, so, letting the first equation be equal to <math>x</math>, we get <math>a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}</math>. Calculating <math>49^2</math> by sheer multiplication is not difficult, but you can also do <math>(50-1)(50-1)=2500-100+1=2401</math>. We want to find the value of <math>x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225</math>. Since <math>x=\frac{137-2401}{2}</math>, we find <math>x=-1132</math>. <math>-1132+1225=\boxed{93}</math>.
+
Note that each term in the second expression is <math>49</math> greater than the corresponding term, so, letting the first equation be equal to <math>x</math>, we get <math>a_1+a_2+a_3+\ldots+a_{98}=137=2x+49^2 \implies x=\frac{137-49^2}{2}</math>. Calculating <math>49^2</math> by sheer multiplication is not difficult, but you can also do <math>(50-1)(50-1)=2500-100+1=2401</math>. We want to find the value of <math>x+\frac{49 \cdot 50}{2}=x+49 \cdot 25=x+1225</math>. Since <math>x=\frac{137-2401}{2}</math>, we find <math>x=-1132</math>. <math>-1132+1225=\boxed{93}.</math>
 
- PhunsukhWangdu
 
- PhunsukhWangdu
  

Revision as of 14:29, 10 August 2021

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

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

Or, since the sum of the odd elements if 44, then the sum of the even terms must be $\fbox{093}$.

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

See also

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
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_1&oldid=159993"