JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    If\[{{\left( 2+\frac{x}{3} \right)}^{55}}\]is expanded in the ascending powers of x and the coefficients of powers of x in two consecutive terms of the expansion are equal, then these terms are:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) 7th and 8th           

    B) 8th and 9th

    C)  28th  and 29th     

    D) 27th  and 28th

    Correct Answer: A

    Solution :

    Let \[{{\text{r}}^{\text{th}}}\]and \[\text{(r+1)}{{\,}^{\text{th}}}\]term has equal coefficient\[{{\left( 2+\frac{x}{3} \right)}^{55}}={{2}^{55}}{{\left( 1+\frac{x}{6} \right)}^{55}}\] \[{{\text{r}}^{\text{th}}}\]term\[={{2}^{55}}{{\,}^{55}}{{C}_{r}}{{\left( \frac{x}{6} \right)}^{r}}\] Coefficient of \[{{x}^{r}}\] is\[{{2}^{55}}{{\,}^{55}}{{C}_{r}}\frac{1}{{{6}^{r}}}\] Coefficient of \[{{x}^{r+1}}\]is\[{{2}^{55}}{{\,}^{55}}{{C}_{r+1}}.\frac{1}{{{6}^{r+1}}}\] \[{{2}^{55}}{{\,}^{55}}{{C}_{r}}\frac{1}{{{6}^{r}}}={{2}^{55}}\,{{\,}^{55}}{{C}_{r+1}}\frac{1}{{{6}^{r+1}}}\] \[\frac{1}{\underline{|r|55-r}}=\frac{1}{\underline{|r+1|54-r}}.\frac{1}{6}\] \[6(r+1)=55-r\] \[6r+6=55-r\] \[7r=49\] \[r=7\] \[(r+1)=9\] Coefficient of 7th and 8th terms are equal.

You need to login to perform this action.
You will be redirected in 3 sec spinner