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 ${(\frac{a}{3} + 3)}^{8}$ is 1120, then value of $a$ is

A
$\pm 2$
B
$\pm 1$
C
$\pm \sqrt{3}$
D
$\pm \sqrt{2}$

Solution:

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

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

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

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

Related content

NCERT Books for Class 10- Download Free PDF (2023-2024)

NCERT Books for Class 11- Download Free PDF (2023-2024)

USA Full Form – United States of America

NRC Full Form – National Register of Citizens

Distance Speed Time Formula

Refractive Index Formula

Electric Current Formula

Ohm’s Law Formula

Wavelength Formula