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**