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Difference between revisions of "2002 AMC 10B Problems/Problem 19"

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<math> \mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1 </math>
 
<math> \mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1 </math>
  
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== Solution ==
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== Solution 1 ==
 +
 
 +
 
 +
We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...
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<cmath>(a_{101} - a_1) + (a_{102} - a_2) +... + (a_{200} - a_{100}) = 200-100 = 100 </cmath>
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...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 ...
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<math>\frac{100}{100} = 1 </math>
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...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...
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<math>\frac{1}{100} =\boxed{(\textbf{C})\ 0.01}</math>
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== Solution 2 ==
 
Adding the two given equations together gives  
 
Adding the two given equations together gives  
  
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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  
  
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<math> \frac{n}{2}(2a_1+d(n-1)) </math>,
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<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  
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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>.
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Subtracting '''(1)''' from '''(2)''' eliminates <math> a_1 </math>, yielding <math> 100d=1 </math>, and <math> d=a_2-a_1=\frac{1}{100}=.01, \boxed{\text{C}} </math>.
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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>.
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 +
== Solution 3 ==
 +
Subtracting the 2 given equations yields
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<math>(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100</math>
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 +
 
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Now express each <math>a_n</math> in terms of first term <math>a_1</math> and common difference <math>x</math> between consecutive terms
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<math>((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100</math>
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 +
 
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Simplifying and canceling <math>a_1</math> and <math>x</math> terms gives
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<math>100x+100x+100x+...+100x=100</math>
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<math>100x\times100=100</math>
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<math>100x=1</math>
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<math>x=0.01=\boxed{(\textbf{C})\ 0.01}</math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/tKsYSBdeVuw?t=4410
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 +
~ pi_is_3.14
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==Video Solution==
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https://www.youtube.com/watch?v=38p1OD_ATFE  ~David
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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
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All AMC 10 Problems and Solutions

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