If the $17^{\text{th}}$ term of an A.P. exceeds its $10^{\text{th}}$ term by $7$, the common difference is:

The $n^{\text{th}}$ term of an Arithmetic Progression is given by $a_n=a+(n-1)d$, where $a$ is the first term and $d$ is the common difference.
According to the question:

$$a_{17}-a_{10}=7$$

$$[a+(17-1)d]-[a+(10-1)d]=7$$

$$(a+16d)-(a+9d)=7$$

$$7d=7$$

$$d=1$$

Therefore, the common difference is $1$.