# If $p$ is a real number and if the middle term in the expansion of $(\large\frac{p}{2}$$+2)^8 is 1120.Find p \begin{array}{1 1}(A)\;\pm 3\\(B)\;\pm 2\\(C)\;\pm 4\\(D)\;\pm 5\end{array} ## 1 Answer Toolbox: • If n is even then the total number of terms in the expansion of (a+b)^n is n+1.Hence there is only one middle term i.e \big(\large\frac{n}{2}$$+1\big)^{th}$ term
• $T_{r+1}=nC_r(a)^{n-r} b^r$
Number of terms in the expansion is $n+1=8+1=9$
Hence the middle term is $\large\frac{n+2}{2}=\frac{8+2}{2}$$=5^{th} T_{r+1}=T_5=8C_4 (\large\frac{p}{2})^4$$2^4$
$\Rightarrow \large\frac{8!}{4!4!}.\frac{p^4}{16}$$.16 \Rightarrow \large\frac{8\times 7\times 6\times 5\times 4!}{4\times 3\times 2\times4!}.\frac{p^4}{16}$$.16$
$\Rightarrow 70p^4$
The middle term is equal to 1120
$\Rightarrow 70p^4=1120$
$\Rightarrow p^4=16$
$\Rightarrow p=\pm 2$
Hence (B) is the correct answer.