Sequences and series
- Thread starter udaisy96
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It's not perfectly clear what the series is, but with a simple assumption we can do something with it.
9 - 5 = 4
13 - 9 = 4
(401 - 13)/4 = 97
Looks like we're good. It's an arithmetic series with a starting value of 5 and a common difference of 4.
Let's see what you get.
9 - 5 = 4
13 - 9 = 4
(401 - 13)/4 = 97
Looks like we're good. It's an arithmetic series with a starting value of 5 and a common difference of 4.
Let's see what you get.
HallsofIvy
Elite Member
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- Jan 27, 2012
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The crucial point of an arithmetic sequence is that the average of all the numbers in the sequence is exactly the same as the sequence of the first and last numbers in the sequence. Here, the first number is 5 and the last number is 401. That means that the average number is \(\displaystyle \frac{5+ 401}{2}= \frac{406}{2}= 203\).
So the question now is "how many numbers are there in this sequence". The constant difference is 9- 5= 4, of course, and there are \(\displaystyle \frac{401- 5}{4}= \frac{396}{4}= 99\); 99+1= 100 terms in the sequence.
So the question now is "how many numbers are there in this sequence". The constant difference is 9- 5= 4, of course, and there are \(\displaystyle \frac{401- 5}{4}= \frac{396}{4}= 99\); 99+1= 100 terms in the sequence.