Find the 4th term from the end in the expansion of

To Find : $4^{\text {th }}$ term from the end

Formulae :

$\cdot \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$\cdot\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)$

For $\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}$

$\mathrm{a}=\frac{4 \mathrm{x}}{5} \mathrm{~b}=\frac{-5}{2 \mathrm{x}}$ and $\mathrm{n}=9$

As n=9, therefore there will be total (9+1)=10 terms in the expansion

Therefore,

$4^{\text {th }}$ term from the end $=(10-4+1)^{\text {th, }}$ i.e. $7^{\text {th }}$ term from the starting.

We have a formula

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

For $t_{7}, r=6$

$\therefore \mathrm{t}_{7}=\mathrm{t}_{6+1}$

$=\left(\begin{array}{c}10 \\ 6\end{array}\right)\left(\frac{4 x}{5}\right)^{10-6}\left(\frac{-5}{2 x}\right)^{6}$

$\left.=\left(\begin{array}{c}10 \\ 4\end{array}\right)\left(\frac{4 \mathrm{x}}{5}\right)^{4}\left(\frac{-5}{2 \mathrm{x}}\right)^{6} \ldots \ldots \ldots\left(\begin{array}{c}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

$=\left(\begin{array}{c}10 \\ 4\end{array}\right) \frac{(4)^{4}}{(5)^{4}}(\mathrm{x})^{4} \frac{(-5)^{6}}{(2)^{6}}(\mathrm{x})^{-6}$

$=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}(100)(\mathrm{x})^{-2}$

$=21000(\mathrm{x})^{-2}$

Therefore, a $4^{\text {th }}$ term from the end $=21000(x)^{-2}$

$\underline{\text { Conclusion : }} 4^{\text {th }}$ term from the end $=21000(\mathrm{x})^{-2}$