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 Question from Laura, a student: In an arithmetic series 5+9+13+...+tn has a sum of 945. How many terms does the series have? What formula do I use?

Hi Laura,

I don't remember many mathematical formulas, especially when it comes to arithmetic series. I'll show you how I would approach this problem.

I can see from the sequence 5, 9, 13, ... that each term is 4 more than the previous term.

The first term is 5.
The second term is 4 plus 5.
The third term is 2 fours plus 5.

Thus the nth term must be n-1 fours plus 5. That is tn = 4(n - 1) + 5 = 1 + 4n.

Thus the n term series is

S = 5 + 9 + 13 + ... + (1 + 4(n-1)) + (1 + 4n)

You can find the sum of this series by writing it forward and backwards and then adding down

 S = 5 + 9 + ... + 1+4(n-1) + 1+4n S = 1+4n + 1+4(n-1) + ... + 9 + 5 2S = 6+4n + 6+4n + ... + 6+4n + 6+4n

Since there are n terms in the series

2S = n(6 + 4n)

and thus

S = n(6 + 4n)/2 = 945

Solve for n.
Penny

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