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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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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} $

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  • 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.
answered Jun 24, 2014 by sreemathi.v

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