Q43 / 120 · VITEEE 2010 · Mathematics
The numer ically greatest term in the expansion of $(3-5 x)^{11}$ when $x=\frac{1}{5}$, is
- Correct answerA$55 \times 3^{9}$
- B$55 \times 3^{6}$
- C$45 \times 3^{9}$
- D$45 \times 3^{6}$
Solution
$\begin{aligned}(3-5 x)^{11} &=3^{11}\left(1-\frac{5 x}{3}\right)^{11} \\ &=3^{11}\left(1-\frac{5}{3} \cdot \frac{1}{5}\right)^{11} \quad\left\{\because x=\frac{1}{5}\right\} \\ &=3^{11}\left(1-\frac{1}{3}\right)^{11} \end{aligned}$ Now, $r=\frac{|x|(n+1)}{|x|+1}=\frac{\left|-\frac{1}{3}\right|(11+1)}{\left|-\frac{1}{3}\right|+1}=\frac{4}{\frac{4}{3}}$ $\begin{aligned} \Rightarrow r=3
\end{aligned}$
Therefore, $3 \mathrm{rd}\left(T_{3}\right)$ and $(3+1)=4$ th $\left(T_{4}\right)$
terms are numerically greatest in the expansion of $(3-5 x)^{11}$
So, greatest term $=T_{3}$
$$
\begin{array}{l}
=3^{11}\left|{ }^{11} C_{2}(1)^{9}\left(-\frac{1}{3}\right)^{2}\right|=3^{11}\left|\frac{11 \times 10}{1.2 .9}\right| \\
=55 \times 3^{9}
\end{array}
$$
and $T_{4}=3^{11}\left|{ }^{11} C_{3}(1)^{8}\left(-\frac{1}{3}\right)^{3}\right|$
$$
\begin{array}{c}
=3^{11}\left|\frac{11 \times 10 \times 9}{1.2 .3} \cdot\left(-\frac{1}{27}\right)\right|=55 \times 3^{9} \\
\therefore \text { Greatest term (numerically) } \\
=T_{3}=T_{4}=55 \times 3^{9}
\end{array}
$$
Hence, the correct answer is option: 'A'.