Difference between revisions of "2015 AMC 8 Problems/Problem 9"
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===Solution 3=== | ===Solution 3=== | ||
| − | We can easily find out she makes <math>2 | + | We can easily find out she makes <math>2\cdot20-1 = 39</math> widgets on Day <math>20</math>. Then, we make the sum of <math>1,3, 5,......,35,37,39</math> by adding in this way: <math>(1+39)+(3+37)+(5+35)+...+(19+21)</math>, which include <math>10</math> pairs of <math>40</math>. So the sum of <math>1,3,5, ........~39</math> is <math>(40\cdot10)=\boxed{\textbf{(D)}~400}</math> ----LarryFlora |
==See Also== | ==See Also== | ||
Revision as of 21:05, 25 December 2021
Problem
On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working 
days?
Solutions
Solution 1
First, we have to find how many widgets she makes on Day . We can write the linear equation 
to represent this situation. Then, we can plug in for 
: 
-- 
-- 
The sum of 
is 
Solution 2
The sum is just the sum of the first odd integers, which is 
Solution 3
We can easily find out she makes 
widgets on Day . Then, we make the sum of 
by adding in this way: 
, which include 
pairs of 
. So the sum of 
is 
----LarryFlora
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
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