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The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its

A. 24th term

B. 27th term

C. 26th term

D. 25th term

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,n\,{\text{terms}}\left( {{S_n}} \right) = 3{n^2} + 5n \cr & \therefore {\text{Sum}}\,{\text{of}}\,\left( {n - 1} \right)\,{\text{terms}}\,\left( {{S_{n - 1}}} \right) \cr & = 3{\left( {n - 1} \right)^2} + 5\left( {n - 1} \right) \cr & = 3\left( {{n^2} - 2n + 1} \right) + 5n - 5 \cr & = 3{n^2} - 6n + 3 + 5n - 5 \cr & = 3{n^2} - n - 2 \cr & \therefore {n^{th}}\,{\text{term}} = {S_n} - {S_{n - 1}} \cr & \Rightarrow {a_n} = 3{n^2} + 5n - 3{n^2} + n + 2 \cr & \,\,\,\,\,\,\,\,\,\,{a_n} = 6n + 2,\,{\text{But}}\,{a_n} = 164 \cr & \Rightarrow 6n + 2 = 164 \cr & \Rightarrow 6n = 164 - 2 \cr & \Rightarrow 6n = 162 \cr & \therefore n = \frac{{162}}{6} = {27^{th}}\,{\text{term}} \cr} $$

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