A) \[n{{(2c)}^{1/n}}\]
B) \[(n+1){{c}^{1/n}}\]
C) \[2n{{c}^{1/n}}\]
D) \[(n+1){{(2c)}^{1/n}}\]
Correct Answer: A
Solution :
[a] We have \[({{a}_{1}}+{{a}_{2}}+....+{{a}_{n-1}}+2{{a}_{n}})/n\ge {{({{a}_{1}}{{a}_{2}}....{{a}_{n-1}}2{{a}_{n}})}^{1/n}}\] \[\Rightarrow \,\,\,{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......+{{a}_{n-1}}+2{{a}_{n}}\ge n{{(2c)}^{1/n}}\]You need to login to perform this action.
You will be redirected in
3 sec