Posted by Sammy on Thursday, March 15, 2012 at 1:21am.
Write a rule for the nth term of the arithmetic sequence. ( I'm not sure how I'm suppose to write the numbers below a :C so bare with me )
a(2) =  28, a(20) = 52
I used the formula a(n) = a(1) + ( n1)d
a(20) = a(1) + (201)d > 52 = a(1)+19d
a(2) = a(1) + (21)d > 28= a(1) + 1d
And that's where I'm stuck >_< please help. Thanks :)

Algebra 2  Steve, Thursday, March 15, 2012 at 11:41am
just write down both equations, using you known values for a(2) and a(20):
a + 19d = 52
a + d = 28
subtract the equations, and the a goes away!
18d = 80
d = 40/9
so, a = 28  40/9 = 292/9
So, leaving out all the nasty nines, the numerators (multiplied by 9) proceed as follows:
292 252 212 172 132
92 52 12 28 68
108 148 188 228 268
308 348 388 428 468
Note that a(2) = 252/9 = 28
and a(20) = 468/9 = 52