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

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

A
2
B
5
C
7
D
0

Solution:

Middle term of ${(\frac{a}{3} + 3)}^{8}$ is the $(\frac{8}{2} + 1) th$ or

$5^{th}$ term. We have

$T_{5} = T_{4 + 1} =^{8} C_{4} {(\frac{a}{3})}^{4} (3)^{4} = 70 a^{4}$

We are given $T_{5} = 1120 \Rightarrow 70 a^{4} = 1120$

$\Rightarrow a^{4} = 16 = 2^{4} \Rightarrow a = \pm 2$

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