jmc

Denote the first three terms by $a,$ $a+d,$ and $a+2d,$ where $a$ and $d$ are positive integers; then the fourth term is $a+30.$ Since the last three terms form an arithmetic sequence, we have $$(a+d)(a+30)=(a+2d{)}^2,$$or $$a^2+(30+d)a+30d=a^2+4ad+4d^2.$$Solving for $a,$ we get $$a=\frac{4d^2-30d}{30-3d}=\frac{2d(2d-15)}{3(10-d)}.$$Since $a$ is positive, we must have $f(d)=\frac{d(2d-15)}{10-d}>0.$ We construct a sign table for this expression: \begin{array}{c|ccc|c} &$d$ &$2d-15$ &$-d+10$ &$f(d)$ \\ \hline$d<0$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$0<d<\frac{15}{2}$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$\frac{15}{2}<d<10$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$d>10$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]\end{array}Since $d>0,$ we must have $\frac{15}{2}<d<10,$ which only gives two possible integer values for $d,$ namely $8$ and $9.$ For $d=8,$ we get $$a=\frac{2\cdot 8\cdot 1}{3\cdot 2}=\frac{8}{3},$$which is not an integer, so we must have $d=9$ and $$a=\frac{2\cdot 9\cdot 3}{3\cdot 1}=18.$$Then the sum of the four terms is $$a+(a+d)+(a+2d)+(a+30)=18+27+36+48=\boxed{129}.$$