Solve the following

Given:

< an > is an A.P.

$\frac{a_{4}}{a_{7}}=\frac{2}{3}$

$\Rightarrow \frac{a+(4-1) d}{a+(7-1) d}=\frac{2}{3}$

$\Rightarrow \frac{a+3 d}{a+6 d}=\frac{2}{3}$

$\Rightarrow 3(a+3 d)=2(a+6 d)$

$\Rightarrow 3 a+9 d=2 a+12 d$

$\Rightarrow a=3 d \ldots(i)$

$\therefore \frac{a_{6}}{a_{8}}=\frac{a+(6-1) d}{a+(8-1) d}$

$\Rightarrow \frac{a_{6}}{a_{8}}=\frac{a+5 d}{a+7 d}$

$\Rightarrow \frac{a_{6}}{a_{8}}=\frac{3 d+5 d}{3 d+7 d} \quad(\operatorname{From}(\mathrm{i}))$

$\Rightarrow \frac{a_{6}}{a_{8}}=\frac{8 d}{10 d}$

$\Rightarrow \frac{a_{6}}{a_{8}}=\frac{4 d}{5 d}=\frac{4}{5}$