Question

The two consecutive terms in the expansion of ${\left(3+2x\right)}^{74}$ whose coefficients are equal, are

• A. $11,12$
• B. $7,8$
• C. $30,31$
• D. None of these

Answer 1

The correct option is C

$30,31$

Explanation for correct option

Let $T_r$ and $T_{r+1}$ be the two consecutive term

From binomial expansion , the coefficient of the terms are

$T_r=\left(\begin{array}{c}n\\ r-1\end{array}\right){\left(3\right)}^{74-r+1}{\left(2\right)}^{r-1}$

$T_{r+1}=\left(\begin{array}{c}n\\ r\end{array}\right){\left(3\right)}^{74-r}{\left(2\right)}^r$

Given that the coefficient of the two terms are equal .

$\therefore$$T_r=T_{r+1}$

$\Rightarrow$ $\left(\begin{array}{c}n\\ r-1\end{array}\right){\left(3\right)}^{74-r+1}{\left(2\right)}^{r-1}=\left(\begin{array}{c}n\\ r\end{array}\right){\left(3\right)}^{74-r}{\left(2\right)}^r$

$\Rightarrow$$\frac{74!}{\left(r-1\right)!\left(75-r\right)!}{\left(3\right)}^{75-r}{\left(2\right)}^{r-1}=\frac{74!}{\left(r\right)!\left(74-r\right)!}{\left(3\right)}^{74-r}{\left(2\right)}^r$

$\Rightarrow$ $\frac{3}{75-r}=\frac{2}{r}$

$\Rightarrow$ $3r=150-2r$

$\Rightarrow$ $r=30$

Hence, option C is correct.