Question

If $kϵR^{+}$ and the middle term of $(\frac{k}{2}+2{)}^8$ is $1120$, then value of k is

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is B $2$

$KϵR^{+}$

The general term, ${(r+1)}^{th}$ term of the expansion

${(\frac{K}{2}+2)}^8$ is $T_{r+1}=8_{Cr}{\left(\frac{K}{2}\right)}^{8-r}{2}^r$

In the above expansion there are $9$ terms, hence the middle term will be the $5th$ term $(r=4)$

$\therefore$ $T_5=8_{C_4}{\left(\frac{K}{2}\right)}^{8-4}{2}^4$

$=\frac{8!}{4!4!}\frac{K^4}{2^4}.2^4$

$=\frac{8\times 7\times 6\times 5}{4\times 3\times 2}{K}^4$

$=70K^4$

Given,

${70K}^4=1120$

$\Rightarrow K^4=112/7$

$\Rightarrow K^4=16$

$\Rightarrow K^4=2^4$

$\therefore$ $K=2$