2002 AMC 10B Problems/Problem 19


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


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


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



$2a_1+99d=2$. *(1)

For the sum of the equations (shown at the beginning of the solution) we have $n=200$, so



$2a_1+199d=3$ *(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


Now express each $a_n$ in terms of first term $a_1$ and common difference $x$ between consecutive terms


Simplifying and canceling $a_1$ and $x$ terms gives




$x=0.01=\boxed{(\textbf{C})\ 0.01}$

Video Solution by OmegaLearn


~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=38p1OD_ATFE ~David

See Also

2002 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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