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

Revision as of 15:02, 28 December 2011

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

@@ Line 36: / Line 36: @@
 ==See Also==
-{{AMC10 box|year=2002|ab=B|num-b=2|num-a=4}}
+{{AMC10 box|year=2002|ab=B|num-b=18|num-a=20}}
 [[Category:Introductory Algebra Problems]]

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

Page

Toolbox

Search

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

Revision as of 15:02, 28 December 2011

Problem

Solution

See Also