A) 2015
B) 4025
C) 5030
D) 6035
E) 8045
Correct Answer: E
Solution :
Given first four term of an AP \[a,\text{ }9,\text{ }3a-b,\text{ }3a+b\] \[\Rightarrow \] \[2-9=a+3a-b,4a-b=18\] ...(i) and \[2(3a-b)=9+3a+b\] \[\Rightarrow \] \[6a-2b=9+3a+b,3a-3b=9\] \[\Rightarrow \] \[a-b=3\] ...(ii) On solving Eqs. (i) and (ii), \[3a=15\Rightarrow a=5\,and\,b=2\] The series becomes, 5, 9, 13, 17. First term = 5 Common difference = 4 Now, \[{{T}_{2011}}=5(2011+1).4\] \[=5+8040=8045\]You need to login to perform this action.
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