If a is a real number and if the middle term of a3+38 is 1120, then value of a is

# If $a$ is a real number and if the middle term of ${\left(\frac{a}{3}+3\right)}^{8}$ is 1120, then value of $a$ is

1. A

$±2$

2. B

$±1$

3. C

$±\sqrt{3}$

4. D

$±\sqrt{2}$

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### Solution:

Middle term of ${\left(\frac{a}{3}+3\right)}^{8}$ is the $\left(\frac{8}{2}+1\right)$ th or ${5}^{th}$ term. We have

${T}_{5}={T}_{4+1}{=}^{8}{C}_{4}{\left(\frac{a}{3}\right)}^{4}\left(3{\right)}^{4}=70{a}^{4}$

We are given ${T}_{5}=1120⇒70{a}^{4}=1120$

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