Question

Which one is/are numerically greatest term in the expansion of $(3-5x{)}^{15};x=\frac{1}{5}$ ?

• A. $T_2$
• B. $T_3$
• C. $T_4$
• D. $T_5$

Answer 1

Answer: (C), (D)

The correct options are
C $T_4$
D $T_5$

$(3-5x{)}^{15}=3^{15}{[1+(-\frac{5x}{3})]}^{15}$
To find numerically greatest term, $\frac{T_{r+1}}{T_r}\ge 1$
$\Rightarrow T_{r+1}={}^{15}{C}_r{\left(\frac{-5x}{3}\right)}^r$
and, $T_r={}^{15}{C}_{r-1}{\left(\frac{-5x}{3}\right)}^{r-1}$
$\frac{T_{r+1}}{T_r}=\frac{{}^{15}{C}_r{(-\frac{5x}{3})}^r}{{}^{15}{C}_{r-1}{(-\frac{5x}{3})}^{r-1}}$
$=\frac{\frac{15!}{r!(15-r)!}}{\frac{15!}{(r-1)!(16-r)!}}\times \left(\frac{-5x}{3}\right)$
$=\frac{16-r}{r}\times \left(\frac{-5x}{3}\right)$
[To find the numerically greatest term, we will ignore the negative sign in $\left(\frac{-5x}{3}\right)$]
Putting $x=\frac{1}{5}$, we get
$=\frac{16-r}{r}\times \left(\frac{1}{3}\right)$
Now, $\frac{T_{r+1}}{T_r}\ge 1$
$\Rightarrow \frac{16-r}{r}\times \left(\frac{1}{3}\right)\ge 1$
$\Rightarrow \frac{16-r}{3r}\ge 1$
$\Rightarrow r\le 4$
So, numerically greatest terms are $T_4$ and $T_5$.