Question

In the expansion of ${(\frac{3x^2}{5}+\frac{5}{3x^2})}^{10}$ mid term is

• A. $291$
• B. $242$
• C. $252$
• D. $284$

Answer 1

The correct option is C $252$

Given expansion $={(\frac{3x^2}{5}+\frac{5}{3x^2})}^{10}$
On comparing with ${(x+a)}^n$, we get
$x=\frac{3x^2}{5}$, $a=\frac{5}{3x^2}$ and $n=10$
$\because$ Number of terms $=10+1=11$
$\therefore$ Mid term $=\left(\frac{11+1}{2}\right)$th term $=6$th term
Now, $T_6=T_{5+1}{=}^{10}{C}_5{\left[\frac{3x^2}{5}\right]}^{10-5}{\left[\frac{5}{3x^2}\right]}^5$ $[\because T_{r+1}{=}^n{C}_r{x}^{n-r}{a}^r]$
${=}^{10}{C}_5{\left[\frac{3x^2}{5}\right]}^5{\left[\frac{5}{3x^2}\right]}^5{=}^{10}{C}_5=252$.