Difference between revisions of "2002 AMC 10B Problems/Problem 19"
m (→Solution 1) |
m (→Solution 3: $LaTeX$) |
||
Line 61: | Line 61: | ||
− | Now express each a_n in terms of first term a_1 and common difference x between consecutive terms | + | Now express each <math>a_n</math> in terms of first term <math>a_1</math> and common difference <math>x</math> between consecutive terms |
Line 67: | Line 67: | ||
− | Simplifying and canceling a_1 and x terms gives | + | Simplifying and canceling <math>a_1</math> and <math>x</math> terms gives |
Revision as of 14:20, 11 October 2015
Problem
Suppose that is an arithmetic sequence with What is the value of
Solution 1
We should realize that the two equations are 100 terms apart, so by subtracting the two equations in a form like...
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...
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...
Solution 2
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 .
Solution 3
Subtracting the 2 given equations yields
Now express each in terms of first term and common difference between consecutive terms
Simplifying and canceling and terms gives
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.