• # 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.

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