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

Revision as of 20:55, 14 November 2017

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...

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

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

$100x\times100=100$

$100x=1$

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

Revision as of 17:03, 25 December 2016 (view source) Onomatopoeia (talk \| contribs) (→‎Solution 1) ← Older edit		Revision as of 20:55, 14 November 2017 (view source) Exl2015 (talk \| contribs) m (→‎Solution 1) Newer edit →
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−	<math>\frac{1}{100} =\boxed{(\text{C}) .01}</math>	+	<math>\frac{1}{100} =\boxed{(\text{C})0.01}</math>

	== Solution 2 ==		== Solution 2 ==

2002 AMC 10B (Problems • Answer Key • Resources)
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