It is given that ${17}^{th}$ term exceeds its ${10}^{th}$ term by 7
It means $a_{17}$ = $a_{10}$ + 7......(1)
Here, $a_{17}$ is the ${17}^{th}$ term and a₁₀ is the ${10}^{th}$ term of an AP.
Using formula $a_n=a+(n-1)d$   to find $n^{th}$ term of arithmetic progression, we get

$a_{17}$ =$a+\left(16\right)d$  .....(2)

$a_{10}$ = $a+\left(9\right)d$ .....(3)

Putting (2) and (3) in equation (1), we get

$a+16d=a+9d+7$

$\Rightarrow 7d=7$

$\Rightarrow d=\frac{7}{7}=1$