Question

The middle term in the expansion of $({x/2+2)}^8$ is 1120, then $x\epsilon R$ is equal to

• A. $-2$
• B. $3$
• C. $-3$
• D. $2$

Answer 1

Answer: (A), (D)

$2$
$-2$

The middle term will be the 5th term.
Hence,
$T_{4+1}={}^8{C}_4(\frac{x}{2}{)}^4{2}^4$
$=70x^4$
$=1120$ ...(given)
Hence,
$x^4=16$
$x=\pm 2$