Q61 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

The numerically greatest term in the binomial expansion of $(2 x-3 y)^5$ when $x=\frac{3}{2}$ and $y=\frac{2}{3}$ is

  • A
    360
  • Correct answer
    B
    1080
  • C
    720
  • D
    2160

Solution

$(2 x-3 y)^5$ Numerically greatest term $=\frac{(n+1)|\alpha|}{|\alpha|+1}$ where $\alpha=\frac{-3 y}{2 x}, n=5$ $$ \because \quad x=\frac{3}{2}, y=\frac{2}{3} $$ $$ \therefore \frac{x}{y}=\frac{9}{4}, \alpha=\frac{-3}{2} \times \frac{4}{9}=\frac{-2}{3} \text {. } $$ $$ \therefore \quad \frac{(n+1)|\alpha|}{|\alpha|+1}=\frac{6 \times \frac{2}{3}}{\frac{2}{3}+1}=\frac{4}{\frac{5}{3}}=\frac{12}{5} \approx 2.4 $$ which is not integer. $\therefore 3$ rd term is numerically greatest term. $\begin{aligned} \mathrm{T}_3 & ={ }^5 C_2(2 x)^3(-3 y)^2 \\ & =10 \times 8 \times 9 \times\left(\frac{3}{2}\right)^3 \times\left(\frac{2}{3}\right)^2-10 \times 27 \times 4=1080 .\end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions