A) \[4n+3\]
B) \[4n+5\]
C) \[4n+6\]
D) \[4n+7\]
Correct Answer: A
Solution :
Given that \[{{S}_{n}}=2{{n}^{2}}+5n\] Putting\[n=1,\ 2,\ 3,\ ..........,\ {{S}_{1}}=2\times 1+5\times 1=7\ \], \[{{S}_{2}}=2\times 4+10=8+10=18,\ {{S}_{3}}=18+15=33\]. So, \[{{T}_{1}}={{S}_{1}}=a=7,\ {{T}_{2}}={{S}_{2}}-{{S}_{1}}=18-7=11\], \[{{T}_{3}}={{S}_{3}}-{{S}_{2}}=33-18=15\] Therefore series is \[7,\,11,\ 15,\,........\] Now, \[{{n}^{th}}\] term\[=a+(n-1)d=7+(n-1)4=4n+3\]. Aliter: As we know \[{{T}_{n}}={{S}_{n}}-{{S}_{n-1}}\] \[=(2{{n}^{2}}+5n)-\left\{ 2\,{{(n-1)}^{2}}+5\,(n-1) \right\}\] \[=2{{n}^{2}}+5n-2{{n}^{2}}+4n-2-5n+5=4n+3\].
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