Fun problem inside
- Thread starter mathisfun
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You could treat it as an arithmetic series.
\(\displaystyle S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\displaystyle S_n\) is the sum of the first n terms, so you want n=19. a is the first term (the one you're after), d is the incrementing amount (here it is 1).
\(\displaystyle S_n = \frac{n}{2}[2a + (n-1)d]\)
\(\displaystyle S_n\) is the sum of the first n terms, so you want n=19. a is the first term (the one you're after), d is the incrementing amount (here it is 1).
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Or, if you haven't covered sequences and series yet, find the average value of the nineteen numbers by taking the total and dividing by the number of numbers. Then note that, since these numbers are consecutive numbers, the average value will also be the middle value of the list.
Eliz.
Eliz.
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Sounds like I was right: you haven't yet covered sequences and series. Just use the method I outlined, instead. It's meant for the more elementary student.mathisfun said:
(Note: I don't mean "elementary" like "stupid" or anything; only as "hasn't covered as much yet". The method I outlined can be used with nothing more than arithmetic methods and a little care.)
Eliz.
Well, since you're calling it a "fun problem":mathisfun said:
Add from 1 to 19: you get 190
remove 19: 171 left; now 0 to 18
remove 19: 152 left; now -1 to 17
remove 19: 133 left; now -2 to 16
remove 19: 114 left; now -3 to 15
remove 19: 95 left; now -4 to 14 : so 19 consecutive integers are -4 to 14