# 2002 AMC 10B Problems/Problem 19

## 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

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 can find the value of the common difference every hundred terms. But we forgot that it happens hundred times. So we have to divide the answer by hundred... $\frac{100}{100} = 1$

The answer yields us the common difference of every hundred terms. So you have to simply divide the answer by hundred again to find the common difference between one term, therefore... $\frac{1}{100} =\boxed{(\text{C}) .01}$


## Solution 2

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 3

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 $100x+100x+100x+...+100x=100$ $100x\times100=100$ $100x=1$ $x=0.01=\boxed{(\text{C})0.01}$

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