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 get the value of the common difference of every hundred terms hundred times. So we have to divide the answer by 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...

$\frac{1}{100} =\boxed{(\text{C})0.01}$

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


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


$100x\times100=100$


$100x=1$


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

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

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