An AP begins 5, 12, 19, ... and has exactly 31 terms. Find its last term.

$a=5,d=7,n=31$. Last term = $a_{31}=5+30\times 7=5+210=215$. The last term is $a+(n-1)d$, not $a+nd$: using 31$d$ gives 222, using 29$d$ gives 208, and 210 is the 30$d$ part on its own without the first term.