Question

If the middle term in the expansion of ${(\frac{x}{2}+2)}^8$is $1120$; then $x\in R$ is equal to

• A. $-2$
• B. $3$
• C. $-3$
• D. $2$

Answer 1

Answer: (A), (D)

$2$

Here $n$ is even
$\therefore$ Middle term is ${(\frac{n}{2}+1)}^{th}\Rightarrow (4+1{)}^{th},i.e.,T_5$
$\therefore T_5={}^8{C}_4{\left(\frac{x}{2}\right)}^4\cdot 2^4=1120\Rightarrow {}^8{C}_4{x}^4=\frac{8\cdot 7\cdot 6\cdot 5}{1\cdot 2\cdot 3\cdot 4}{x}^4=1120$
$\Rightarrow x^4=\frac{1120}{70}=16$
$\Rightarrow x^4-16=0$
$\Rightarrow (x^2+4)(x^2-4)=0$
$\therefore x=\pm 2$ as $x\in R$