4,9,14,19,....a20,...an
We notice that the differenc ebetween terms is 5
==> r= 5
a1= 4
a2= 4+ 5 = 9
a3= 4+ 2(5) = 14
......
a20 = 4 + 19(5) = 99
an = a1 + (n-1)*r
= 4 + (n-1)*5
= 4+ 5n - 5
= 5n -1
==> an = 5n -1
First we have to find the difference between terms:
an = a1 + (n-1)d
d = (an - a1) / (n -1)
= (9 - 4) / 1
= 5
To test that this is an arithmetic series we can determine whether the differences are the same for the other two terms:
d = (14 - 4) / 2 = 5
d = (19 - 4) / 3 = 5
So it is an arithmetic series. From here solving for a20 is as simple as using the same expression:
a20 = a1 + (20 - 1)d
a20 = 4 + 19(5)
a20 = 99