Difference between revisions of "2002 AMC 10B Problems/Problem 19"
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Now we have a system of equations in terms of <math> a_1 </math> and <math> d </math>. | Now we have a system of equations in terms of <math> a_1 </math> and <math> d </math>. | ||
− | Subtracting '''(1)''' from '''(2)''' eliminates <math> a_1 </math>, yielding <math> 100d=1 </math>, and <math> d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C }) .01} </math>. | + | Subtracting '''(1)''' from '''(2)''' eliminates <math> a_1 </math>, yielding <math> 100d=1 </math>, and <math> d=a_2-a_1=\frac{1}{100}=\boxed{(\text{C}) .01} </math>. |
==See Also== | ==See Also== |
Revision as of 23:27, 17 November 2013
Problem
Suppose that is an arithmetic sequence with What is the value of
Solution
Adding the two given equations together gives
.
Now, let the common difference be . Notice that , so we merely need to find to get the answer. The formula for an arithmetic sum is
,
where is the first term, is the number of terms, and 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 . Therefore, we have
,
or
. *(1)
For the sum of the equations (shown at the beginning of the solution) we have , so
or
*(2)
Now we have a system of equations in terms of and . Subtracting (1) from (2) eliminates , yielding , and .
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 | ||
All AMC 10 Problems and Solutions |
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