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 "2002 AMC 10B Problems/Problem 19"

(→‎Solution 3)
(→‎Solution 2)
 
(4 intermediate revisions by 4 users not shown)
Line 12: Line 12:
 
We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...
 
We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...
  
−
 
+
<cmath>(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100 </cmath>
−
<math>(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100 </math>
 
−
 
 
−
 
 
  
 
...we get the value of the common difference of every hundred terms one hundred times. So we have to divide the answer by one hundred to get ...  
 
...we get the value of the common difference of every hundred terms one hundred times. So we have to divide the answer by one hundred to get ...  
Line 23: Line 20:
 
...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...
 
...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...
  
−
<math>\frac{1}{100} =\boxed{(\text{C})0.01}</math>
+
<math>\frac{1}{100} =\boxed{(\textbf{C})\ 0.01}</math>
  
 
== Solution 2 ==
 
== Solution 2 ==
Line 32: Line 29:
 
Now, let the common difference be <math> d </math>. Notice that <math> a_2-a_1=d </math>, so we merely need to find <math> d </math> to get the answer. The formula for an arithmetic sum is  
 
Now, let the common difference be <math> d </math>. Notice that <math> a_2-a_1=d </math>, so we merely need to find <math> d </math> to get the answer. The formula for an arithmetic sum is  
  
−
<math> \frac{n}{2}(2a_1+d(n-1)) </math>,
+
<math> \frac{n}{2}\left(2a_1+d(n-1)\right) </math>,
  
 
where <math> a_1 </math> is the first term, <math> n </math> is the number of terms, and <math> d </math> is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have <math> n=100 </math>. Therefore, we have  
 
where <math> a_1 </math> is the first term, <math> n </math> is the number of terms, and <math> d </math> is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have <math> n=100 </math>. Therefore, we have  
Line 51: Line 48:
  
 
Now we have a system of equations in terms of <math> a_1 </math> and <math> d </math>.
 
Now we have a system of equations in terms of <math> a_1 </math> and <math> d </math>.
−
Subtracting '''(1)''' from '''(2)''' eliminates <math> a_1 </math>, yielding <math> 100d=1 </math>, and <math> d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C}) .01} </math>.
+
Subtracting '''(1)''' from '''(2)''' eliminates <math> a_1 </math>, yielding <math> 100d=1 </math>, and <math> d=a_2-a_1=\frac{1}{100}=\boxed{(\textbf{C})\ 0.01}</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
Line 77: Line 74:
 
<math>100x=1</math>
 
<math>100x=1</math>
  
−
<math>x=0.01</math>
 
  
−
== Video Solution ==
+
<math>x=0.01=\boxed{(\textbf{C})\ 0.01}</math>
 +
 
 +
== Video Solution by OmegaLearn ==
 
https://youtu.be/tKsYSBdeVuw?t=4410
 
https://youtu.be/tKsYSBdeVuw?t=4410
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
−
 
+
==Video Solution==
−
<math>x=0.01=\boxed{(\text{C})0.01}</math>
+
https://www.youtube.com/watch?v=38p1OD_ATFE  ~David
  
 
==See Also==
 
==See Also==

Latest revision as of 09:21, 22 August 2024

Problem

Suppose that $\{a_n\}$ is an arithmetic sequence with \[a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.\] What is the value of $a_2 - a_1 ?$

$\mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1$

Solution 1

We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...

\[(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100\]

...we get the value of the common difference of every hundred terms one hundred times. So we have to divide the answer by one hundred to get ...

$\frac{100}{100} = 1$

...the common difference of every hundred terms. Then we have to simply divide the answer by hundred again to find the common difference between one term, therefore...

$\frac{1}{100} =\boxed{(\textbf{C})\ 0.01}$

Solution 2

Adding the two given equations together gives

$a_1+a_2+...+a_{200}=300$.

Now, let the common difference be $d$. Notice that $a_2-a_1=d$, so we merely need to find $d$ to get the answer. The formula for an arithmetic sum is

$\frac{n}{2}\left(2a_1+d(n-1)\right)$,

where $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have $n=100$. Therefore, we have

$50(2a_1+99d)=100$,

or

$2a_1+99d=2$. *(1)

For the sum of the equations (shown at the beginning of the solution) we have $n=200$, so

$100(2a_1+199d)=300$

or

$2a_1+199d=3$ *(2)

Now we have a system of equations in terms of $a_1$ and $d$. Subtracting (1) from (2) eliminates $a_1$, yielding $100d=1$, and $d=a_2-a_1=\frac{1}{100}=\boxed{(\textbf{C})\ 0.01}$.

Solution 3

Subtracting the 2 given equations yields


$(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100$


Now express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms


$((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100$


Simplifying and canceling $a_1$ and $x$ terms gives


$100x+100x+100x+...+100x=100$


$100x\times100=100$


$100x=1$


$x=0.01=\boxed{(\textbf{C})\ 0.01}$

Video Solution by OmegaLearn

https://youtu.be/tKsYSBdeVuw?t=4410

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=38p1OD_ATFE ~David

See Also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_19&oldid=225547"