Difference between revisions of "2002 AMC 10B Problems/Problem 19"
NumberNinja (talk | contribs) (→Solution) |
NumberNinja (talk | contribs) (→Solution 2) |
||
Line 41: | Line 41: | ||
Now express each a_n in terms of first term a_1 and common difference x between consecutive terms | Now express each a_n in terms of first term a_1 and common difference x between consecutive terms | ||
+ | |||
<math>((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100</math> | <math>((a_1+100x)-(a_1))+((a_1+101x)-(a_1+x))+((a_1+102x)-(a_1+2x))+...+((a_1+199x)-(a_1+99x))=100</math> | ||
+ | |||
Simplifying and canceling a_1 and x terms gives | Simplifying and canceling a_1 and x terms gives | ||
+ | |||
<math>100x+100x+100x+...+100x=100, 100x\times(100)=100, 100x=1, x=0.01</math> | <math>100x+100x+100x+...+100x=100, 100x\times(100)=100, 100x=1, x=0.01</math> |
Revision as of 17:28, 26 November 2014
Contents
Problem
Suppose that is an arithmetic sequence with What is the value of
Solution 1
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 2
Subtracting the 2 given equations yields
Now express each a_n in terms of first term a_1 and common difference x between consecutive terms
Simplifying and canceling a_1 and x 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.