Problem 79

Given that angles $A$, $B$, $C$ form an arithmetic sequence,
$\therefore 2B=A+C$,
Since the sum of angles in a triangle is $\pi$,
$\therefore B= \frac{ \pi}{3}$,
Given $S_{\triangle ABC}= \frac{1}{2}ac \sin B= \frac{ \sqrt{3}}{2}$, $a=1$, $\sin B= \frac{ \sqrt{3}}{2}$,
$\therefore c=2$,
By applying the cosine rule, we get: $b^{2}=a^{2}+c^{2}-2ac \cos B=1+4-2=3$,
Hence, $b= \sqrt{3}$.

So the answer is: $\boxed{b = \sqrt{3}}$.

The problem is solved by utilizing the properties of an arithmetic sequence and the sum of angles in a triangle to find the measure of angle $B$. Then, the formula for the area of a triangle is used along with the given information to find the value of $c$. Finally, the cosine rule is applied to find the value of $b$.

This problem tests your understanding of the cosine rule, the formula for the area of a triangle, and the properties of arithmetic sequences. Proficiency in these theorems and formulas is essential for solving this problem.