The first four terms of an AP are a, 9, `3a- b, 3a+b`.
The 2011 th term of an AP is

Options

A. 2015
B. 4025
C. 5030
D. 8045

Answer

D

Explanation

Given first four terms of an AP a, 9, `3a-b, 3a+b`
`implies2.9=a+3a-bimplies4a-b=18" "…(i)`
and `2(3a-b)=9+3a+b`
`implies" "6a-2b=9+3a+b`
`implies" "3a-3b=9`
`implies" "a-b=3" "…(ii)`
On solving Eqs. (i) and (ii), we get
`3a=15impliesa=5`
and b = 2
The series becomes, 5, 9, 13, 17.
First term = 5,
Common difference = 4
Now, `T_(2011)=5+(2011-1).4`
`=8045`