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

m (Solution)
m (Solution)
Line 33: Line 33:
  
 
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>.
  
 
==See Also==
 
==See Also==

Revision as of 23:27, 17 November 2013

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

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

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS