Question

If $k\in R$ and the middle term of ${\left(\frac{k}{2}+2\right)}^8$ is $1120$ , then value of $k$ is:

• A. 3
• B. 2
• C. $-3$
• D. - 4

Answer 1

Answer: (B)

middle term $=T_5$
$T_5=T_{4+1}={}^8{C}_4\cdot k^4=1120$
$\Rightarrow k=2$