In an AP, it is being given that

To Find: $\frac{T 7}{T 10}$

Given: $\frac{T_{4}}{T_{7}}=\frac{2}{3}$

(Where $T_{n}$ is nth term and $d$ is common difference of given AP)

Formula Used: $T_{n}=a+(n-1) d$

$\frac{\mathrm{T}_{4}}{\mathrm{~T}_{7}}=\frac{2}{3} \rightarrow \frac{a+3 d}{a+6 d}=\frac{2}{3}$ (cross multiply)

$3 a+9 d=2 a+12 d \rightarrow a=3 d$.......equation (i)

Now $\frac{\mathrm{T}_{7}}{\mathrm{~T}_{10}}=\frac{a+6 d}{a+9 d} \rightarrow \frac{\mathrm{T}_{7}}{\mathrm{~T}_{10}}=\frac{3 d+6 d}{3 d+9 d}=\frac{9 d}{12 d}$

$\frac{\mathrm{T}_{7}}{\mathrm{~T}_{10}}=\frac{3}{4}$

So $\frac{\mathrm{T}_{7}}{\mathrm{~T}_{10}}=\frac{3}{4}$