Question

If the coefficient of $5^{\text{th }}$ and $3^{\text{rd }}$ terms in the expansion of ${\left(\sqrt{\frac{15}{2}}x+\frac{1}{x^2}\right)}^n$ where $n\in N$ are equal, then the term independent of $x$ is

• A. $5^{\text{th }}$ term
• B. $4^{\text{th }}$ term
• C. $6^{\text{th }}$ term
• D. $7^{\text{th }}$ term

Answer 1

Answer: (A)

$T_{r+1}={}^n{C}_r{\left(\frac{15}{2}\right)}^{\frac{n-r}{2}}\cdot (x{)}^{-2r}\cdot (x{)}^{n-r}$
$\therefore {}^n{C}_4{\left(\frac{15}{2}\right)}^{\frac{n-4}{2}}={}^n{C}_2{\left(\frac{15}{2}\right)}^{\frac{n-2}{2}}\Rightarrow n=12$
$\therefore T_{r+1}={}^{12}{C}_r{\left(\frac{15}{2}\right)}^{\frac{12-r}{2}}\cdot (x{)}^{12-3r}$
Put $r=4\Rightarrow T_5$ is independent of $x$