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

Explanation

We have,

`(12−9)=(15−12)=(18−15)=3`

Therefore, the given sequence is an A.P. with common difference 3.

a=First term =9

∴`16^(th) ` term `=a _16 =a+(16−1)d=a+15d [∵a_n =a+(n−1)d]`

⟹ `a_16``=9+15*3=54`

`∵` General term `=``(n ^(th)`term `=a+(n−1)d`
∴`a _n` `=9+(n−1)×3=3n+6`