Question

If the middle term in the expansion of ${(\frac{p}{2}+2)}^8$ is 1120, find p.

Answer 1

$T_{r+1}in{(\frac{p}{2}+2)}^8{=}^8{C}_r{\left(\frac{p}{2}\right)}^{8-r}{.2}^r.$

$\text{Middle term}$ $=(\frac{8}{2}+1)th$ $=\text{5th term}$ ${=}^8{C}_4{\left(\frac{p}{2}\right)}^4\times 2^4$

$\therefore \frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}\times p^4=1120\Rightarrow p^4=16\Rightarrow p=\pm 2$