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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 ==
 
Adding the two given equations together gives  
 
Adding the two given equations together gives  
  
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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>.
 
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{(\text{C}) .01} </math>.
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 +
== Solution 2 ==
 +
Subtracting the 2 given equations yields
 +
 +
<math>(a_{101}-a_1)+(a_{102}-a_2)+(a_{103}-a_3)+...+(a_{200}-a_{100})=100</math>
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 +
Now express each a_n in terms of first term a_1 and common difference x 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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 +
Simplifying and canceling a_1 and x terms gives
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 +
<math>100x+100x+100x+...+100x=100, 100x\times(100)=100, 100x=1, x=0.01</math>
  
 
==See Also==
 
==See Also==

Revision as of 18:27, 26 November 2014

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

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

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

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 $d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C}) .01}$.

Solution 2

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

$100x+100x+100x+...+100x=100, 100x\times(100)=100, 100x=1, x=0.01$

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

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