Question

The numerically greatest term in the expansion of $(3-5x)^{11}$ when $x = \frac{1}{5},$ is

• A. $55 \times 3^9 $
• B. $55 \times 3^6$
• C. $45 \times 3^9 $
• D. $45 \times 3^6$

Answer 1

The Correct Option is A

we have
$\left(3-5\right)^{11} = 3^{11} \left(1 - \frac{5x}{3}\right)^{11} $
$ = 3^{11} \left(1- \frac{5}{3}. \frac{1}{5}\right)^{11} \left\{\because x = \frac{1}{5}\right\} $
$= 3^{11} \left(1- \frac{1}{3}\right)^{11} $
Now, $ r= \frac{\left|x\right|\left(n+1\right)}{\left|x\right|+1}$
$= \frac{\left|- \frac{1}{3}\right|\left(11+1\right)}{\left|- \frac{1}{3}\right| + 1}$
$= \frac{4}{\frac{4}{3}} $
$ \Rightarrow r = 3 $
Therefore, $3rd (T_3)$ and $(3 + 1) = 4th (T_4)$ terms are numerically greatest in the expansion of $(3 - 5x)^{11}$.
Hence, greatest term = $T_3$
$ =3^{11} \left|^{11}C_{2} \left(1\right)^{9} \left(- \frac{1}{3}\right)^{2} \right|$
$ =3^{11} \left|\frac{11\times10}{1.2.9}\right| $
$ = 55 \times3^{9} $
and $ T_{4} =3^{11}\left|^{11}C_{3} \left(1\right)^{8} \left(- \frac{1}{3}\right)^{3}\right| $
$ = 3^{11} \left|\frac{11 \times10\times9}{1.2.3} .\left( - \frac{1}{27}\right)\right| = 55\times3^{9} $
Hence Greatest term (numerically) $ = T_{3} =T_{4}$
$= 55 \times3^{9} $