SOLUTION: The arithmetic series 3+7+11+15+...is given. How many terms have a sum less than 500.

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Question 786119: The arithmetic series 3+7+11+15+...is given.
How many terms have a sum less than 500.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
7-3 = 4, 11=7 = 4, 15 - 11 = 4, so common difference -= d = 4

Sn = n%2F2[2a1 + (n-1)d]

where a1 = first term = 3.

Sn = n%2F2[2a1 + (n-1)d] < 500

                n%2F2[2(3) + (n-1)(4)] < 500

Multiply both sides by 2 to get rid of the fraction:

                 n[2(3) + (n-1)(4)] < 1000
             
                      n[6 + 4(n-1)] < 1000

                      n[6 + 4n - 4] < 1000

                          n[2 + 4n] < 1000

                           2n + 4nē < 1000

Divide through by 2

                            n + 2nē < 500  

Get 0 on the right and arrange terms on the left in
descending order:

                      2nē + n - 500 < 0 

Using the quadratic formula we see that this has critical 
values approximately 15.6 and -16.1

We ignore the negative critical value and take the largest integer 
that does not exceed 15.1, which is 15.

So the first 15 terms have a sum less than 500.

answer = 15

Checking:

3+7+11+15+19+23+27+31+35+39+43+47+51+55+59 = 465

The 16th term would be 63, which would give a sum 528, which is over 500,
so

answer = 15 is true. 

Edwin