Question 1040154
<pre>
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[n]}}}{{{""=""}}}{{{expr(n/2)(a[1]+(n+1)d)}}}

and learn what the terms mean:

n = the number of terms
S<sub>n</sub> = the sum of the first n terms
a<sub>1</sub> = the first term
d = the common difference

We substitute S<sub>n</sub>=230, a<sub>1</sub>=5, and d=4
in 

{{{S[n]}}}{{{""=""}}}{{{expr(n/2)(2a[1]+(n-1)d)}}}

and get

{{{230}}}{{{""=""}}}{{{expr(n/2)(2*5+(n-1)4)}}}

Multiply 2*5 and get 10

{{{230}}}{{{""=""}}}{{{expr(n/2)(10+(n-1)4)}}}

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

{{{230}}}{{{""=""}}}{{{expr(n/2)(10+4(n-1))}}}

Distribute the 4 into the (n-1)

{{{230}}}{{{""=""}}}{{{expr(n/2)(10+4n-4)}}}

Combine the 10 and the -4 getting 6 

{{{230}}}{{{""=""}}}{{{expr(n/2)(6+4n)}}}

Clear the fraction by multiplying both sides by 2:

{{{2*230}}}{{{""=""}}}{{{2*expr(n/2)(6+4n)}}}

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

{{{460}}}{{{""=""}}}{{{cross(2)*expr(n/cross(2))(6+4n)}}}

Now we have:

{{{460}}}{{{""=""}}}{{{n(6+4n)}}}

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</pre>