RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    In the expansion of\[{{\left( \frac{2{{x}^{2}}}{3}+\frac{3}{2{{x}^{2}}} \right)}^{10}},\]the middle term is

    A)  251            

    B)  252

    C)  250             

    D)  None of these

    Correct Answer: B

    Solution :

     Mid term in the expansion of\[{{\left( \frac{2{{x}^{2}}}{3}+\frac{3}{2{{x}^{2}}} \right)}^{10}}\]is \[{{T}_{\frac{10}{2}+1}}{{=}^{10}}{{C}_{\frac{10}{2}}}{{\left( \frac{2{{x}^{2}}}{3} \right)}^{\frac{10}{2}}}{{\left( \frac{3}{2{{x}^{2}}} \right)}^{\frac{10}{2}}}\] \[\because \] \[\left[ {{T}_{\frac{n}{2}+1}}{{=}^{n}}{{C}_{\frac{n}{2}}}{{(x)}^{\frac{n}{2}}}{{(a)}^{\frac{n}{2}}} \right]\] \[\Rightarrow \] \[{{T}_{6}}{{=}^{10}}{{C}_{5}}{{\left( \frac{2{{x}^{2}}}{3} \right)}^{5}}{{\left( \frac{3}{2{{x}^{2}}} \right)}^{5}}\] \[\Rightarrow \] \[{{T}_{6}}{{=}^{10}}{{C}_{5}}.1\] \[\Rightarrow \] \[{{T}_{6}}=252\]


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