SOLUTION: How many terms of the series 5+9+13+17... Must be added so that the sum will be 230? Please show the solution

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Question 1040154: How many terms of the series 5+9+13+17... Must be added so that the sum will be 230? Please show the solution
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
First we find the common difference by subtracting
consecutive terms:

9-5=4
13-9=4
17-13=4

So the common difference d=4

Then we learn the formula for the sum of the first
n terms:

S%5Bn%5D%22%22=%22%22expr%28n%2F2%29%28a%5B1%5D%2B%28n%2B1%29d%29

and learn what the terms mean:

n = the number of terms
Sn = the sum of the first n terms
a1 = the first term
d = the common difference

We substitute Sn=230, a1=5, and d=4
in 

S%5Bn%5D%22%22=%22%22expr%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

and get

230%22%22=%22%22expr%28n%2F2%29%282%2A5%2B%28n-1%294%29

Multiply 2*5 and get 10

230%22%22=%22%22expr%28n%2F2%29%2810%2B%28n-1%294%29

Move the 4 in front of the (n-1)

230%22%22=%22%22expr%28n%2F2%29%2810%2B4%28n-1%29%29

Distribute the 4 into the (n-1)

230%22%22=%22%22expr%28n%2F2%29%2810%2B4n-4%29

Combine the 10 and the -4 getting 6 

230%22%22=%22%22expr%28n%2F2%29%286%2B4n%29

Clear the fraction by multiplying both sides by 2:

2%2A230%22%22=%22%222%2Aexpr%28n%2F2%29%286%2B4n%29

Multiply the 2 times the 230 on the left getting 460.
Cancel the 2's on the right:

460%22%22=%22%22cross%282%29%2Aexpr%28n%2Fcross%282%29%29%286%2B4n%29

Now we have:

460%22%22=%22%22n%286%2B4n%29

Can you solve that for n?  If not ask me how in the thank-you
note form below and I'll get back to you by email.

Edwin

Answer by ikleyn(52788) About Me  (Show Source):