A) \[\frac{17}{2}\]
B) \[\frac{15}{2}\]
C) 7
D) 8
Correct Answer: C
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{\frac{1}{(a+1)}\,.{{n}^{a+1}}+{{a}_{1}}{{n}^{a}}+{{a}_{2}}{{n}^{a-1}}+.....}{{{(n+1)}^{a-1}}.{{n}^{2}}\left( a+\frac{1+\frac{1}{n}}{2} \right)}=\frac{1}{60}\] \[\Rightarrow \,\,\,\,\,\,\,\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{\frac{1}{a+1}\,+\frac{{{a}_{1}}}{n}+\frac{{{a}_{2}}}{{{n}^{2}}}+....}{{{\left( 1+\frac{1}{n} \right)}^{a}}\left( a+\frac{1+\frac{1}{n}}{2} \right)}=\frac{1}{60}\] \[\Rightarrow \] \[\frac{\frac{1}{a+1}}{\left( a+\frac{1}{2} \right)}\,=\frac{1}{60}\Rightarrow \,(a+1)\,(2a+1)\,=120\] \[2{{a}^{2}}+3a-119=0\] \[2{{a}^{2}}+17a-14a-119=0\] \[\Rightarrow \] \[(a-7)\,(2a+17)=0\] \[a=7,\,\,-\,\frac{17}{2}\]
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