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
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) + ( n-1)d
a(20) = a(1) + (20-1)d > 52 = a(1)+19d
a(2) = a(1) + (2-1)d > -28= a(1) + 1d
And that's where I'm stuck >_< please help. Thanks :)
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