Question

The numerically greatest term in the expansion of $(2x-3y{)}^{12}$) when $x=1,y=\frac{5}{3}$, is

• A. 9^th
• B. 10^th
• C. 11^th
• D. 12^th

Answer 1

Answer: (B)

Given expansion is $(2x-3y{)}^{12}$

Since, Numerically greatest term in the expansion $(1-\alpha {)}^n$ is

$\frac{(n+1)|\alpha |}{|\alpha |+1}$ th term.

Now,

$(2x-3y{)}^{12}$

$=\left(2x{)}^{12}\right(1-\frac{3y}{2x}{)}^{12}$

Here, $|\alpha |=|-\frac{3y}{2x}|,n=12$

Numerically greatest term at $x=1,y=\frac{5}{3}$ is,

$\frac{(n+1)|\alpha |}{|\alpha |+1}$ th term.

$=\frac{(12+1)|-\frac{3y}{2x}|}{|-\frac{3y}{2x}|+1}$

$=\frac{(12+1)|\frac{3y}{2x}|}{|\frac{3y}{2x}|+1}$

$=\frac{(12+1)|\frac{3\left(\frac{5}{3}\right)}{2\left(1\right)}|}{|\frac{3\left(\frac{5}{3}\right)}{2\left(1\right)}|+1}$

$=\frac{13\left(\frac{5}{2}\right)}{\frac{5}{2}+1}$

$=\frac{\frac{65}{2}}{\frac{7}{2}}$

$=\frac{65}{2}\times \frac{2}{7}$

$=9.28$

Since, $9<9.28<10$

Hence, the numerically greatest term is ${10}^{\text{th}}$ term