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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
$8^{th}$ and $9^{th}$
$7^{th}$ and $8^{th}$
$28^{th}$ and $29^{th}$
$27^{th}$ and $28^{th}$
Solution
Let $(r+1)^{\text {th }}$ and $(r+2)^{\text {th }}$ term has equal coefficient
$\left(2+\frac{x}{3}\right)^{55}=2^{55}\left(1+\frac{x}{6}\right)^{55}$
$(r+1)^{\text {th }}$ term $=2^{55\, 55} \mathrm{C}_{r}\left(\frac{x}{6}\right)^{r}$
Coefficient of $x^{r}$ is $2^{55\, 55} \mathrm{C}_{r} \frac{1}{6^{r}}$
$(r+2)^{t h}$ term $=2^{55\,55} C_{r+1}\left(\frac{x}{6}\right)^{r+1}$
Coefficient of $x^{r+1}$ is $2^{55\, 55} \mathrm{C}_{r+1} \cdot \frac{1}{6^{r+1}}$
Both coefficients are equal
$2^{55\,55} C_{r} \frac{1}{6^{r}}=2^{55\, 55} \mathrm{C}_{r+1} \frac{1}{6^{r+1}}$
$\frac{1}{{r!55 – r!}} = \frac{1}{{r – 1!54 – r!}}.\frac{1}{6}$
$6(r+1)=55-r$
$6 r+6=55-r$
$7 r=49$
$r=7$
$(r+1)=8$
Coefficient of $8^{th}$ and $9^{th}$ terms are equal.