A) 14
B) 35
C) 80
D) 40
Correct Answer: D
Solution :
Here\[{{T}_{n}}=3n-1\], putting \[n=1,\ 2,\ 3,\ 4,\ 5\] we get first five terms, \[2,\ 5,\ 8,\ 11,\ 14\] Hence sum is\[{{n}^{th}}\]. Aliter: \[{{S}_{n}}=\Sigma {{T}_{n}}=3\Sigma n-\Sigma 1=\frac{3n(n+1)}{2}-n\] \[\Rightarrow \] \[{{S}_{5}}=\frac{3.5.6}{2}-5=40\].You need to login to perform this action.
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