Question

The middle term in the expansion ${(\frac{x^2}{4}-\frac{4}{x^2})}^n$ is $-252$.
Find the third term from end.

• A. ${45.2}^6{x}^{-12}$
• B. ${25.4}^6{x}^{12}$
• C. ${45.4}^6{x}^{-12}$
• D. ${45.4}^{-6}{x}^{-12}$

Answer 1

Answer: (C)

${45.4}^6{x}^{-12}$

Given the middle term in the expansion of $(\frac{x^2}{4}-\frac{4}{x^2}{)}^n$ is $-252$.

Let $n$ be odd number,then there exists two middle terms however it will not be a constant term.

So $n$ must be an even number.

we k.n.t $T_r+1$ in expansion of $(a+b{)}^n$ is ${}^n{C}_r{a}^{n-r}{b}^r$

$\therefore T_{\frac{n}{2}+1}{=}^n{C}_{\frac{n}{2}}.(\frac{x^2}{4}{)}^{\frac{n}{2}}.(-\frac{4}{x^2}{)}^{\frac{n}{2}}=(-1{)}^{\frac{n}{2}}{.}^n{C}_{\frac{n}{2}}$

$(-1{)}^{\frac{n}{2}}{.}^n{C}_{\frac{n}{2}}=-252$

Thus $\frac{n}{2}$ is an odd number.

${}^n{C}_{\frac{n}{2}}=252$

$\Rightarrow n=10$

Third term from end = (10-3+2) term from begining

$\therefore T_9{=}^{10}{C}_8.(\frac{x^2}{4}{)}^{10-8}.(-\frac{4}{x^2}{)}^8=$${}^{10}{C}_2.(-1{)}^8.(\frac{x^2}{4}{)}^{2-8}=45.x^{-12}{.4}^6$

$\therefore$Third term from end is ${45.4}^6{x}^{-12}$