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

Revision as of 13:39, 8 January 2015

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

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

$100x\times100=100$

$100x=1$

$x=0.01=\boxed{(C)\0.01}$ (Error compiling LaTeX. Unknown error_msg)

Solution 3

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

$(a_101 - a_1) + (a_102 - a_2) + (a_103 - a_3) + ... = 200-100 = 100$

We can find the value of the common difference every hundred terms. But we forgot that it happens hundred times. So we have a divide the answer by hundred

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

The answer yields us the common difference of every hundred terms. So you has to simply divide the answer by hundred again to find the common difference between one term

@@ Line 61: / Line 61: @@
 <math>x=0.01=\boxed{(C)\0.01}</math>
+== Solution 3 ==
+We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like
+<math>(a_101 - a_1) + (a_102 - a_2) + (a_103 - a_3) + ... = 200-100 = 100 </math>
+We can find the value of the common difference every hundred terms. But we forgot that it happens hundred times. So we have a divide the answer by hundred
+<math>\frac{100}{100} = 1 </math>
+The answer yields us the common difference of every hundred terms. So you has to simply divide the answer by hundred again to find the common difference between one term
+ <math>\frac{1}{100} =\boxed{(\text{C}) .01}</math>
 ==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
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