Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Review Exercises - Page 839: 21

The common difference is: $d=1.$ The initial term: $a_1=-9.$ $a_n=-10+n$, $a_n=a_{n-1}+1$

We know that $a_{10}=0,a_{18}=8$. Thus the common difference is: $d=\frac{a_k-a_l}{k-l}=\frac{a_{18}-a_{10}}{18-10}=\frac{8-0}{8}=1.$ The initial term: $a_1=a_n-(n-1)d=a_{10}-(9)d=0-(9)1=-9.$ Thus: $a_n=a_1+(n-1)d=-9+(n-1)1=-10+n$, $a_n=a_{n-1}+d=a_{n-1}+1$