Find the 7th term in the expansion of

To find: $7^{\text {th }}$ term in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $T_{r+1}={ }^{n} C_{r} a^{n-r} b^{r}$

For $7^{\text {th }}$ term, $\mathrm{r}+1=7$

$\Rightarrow r=6$

$\ln \left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$

$7^{\text {th }}$ term $=\mathrm{T}_{6+1}$

$\Rightarrow{ }^{8} \mathrm{C}_{6}\left(\frac{4 \mathrm{x}}{5}\right)^{8-6}\left(\frac{5}{2 \mathrm{x}}\right)^{6}$

$\Rightarrow \frac{8 !}{6 !(8-6) !}\left(\frac{4 x}{5}\right)^{2}\left(\frac{5}{2 x}\right)^{6}$

$\Rightarrow(28)\left(\frac{16 x^{2}}{25}\right)\left(\frac{15625}{64 x^{6}}\right)$

$\Rightarrow \frac{4375}{x^{4}}$

Ans) $\frac{4375}{x^{4}}$