Show that the sequence 9,12,15,18,... is an A.P. Find its 16th term and the general term.

Explanation

since (12-9)=(15-12)=(18-15)=3, the given sequence is an A.P. with common difference 3. The first term is 9, therefore, the `16^(th)` term is
`a_(16)=a+(16-1)d` [`because a_(n)=a+(n-1)d`]
=a+15d
`=9+15xx3=54`
The general term (nth term) is given by
`a_(n)=a+(n-1)d`
`=9+(n-1)xx3=3n+6`