[tex]\bf a_n=\cfrac{20}{3}+\cfrac{1}{3}(n-1) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{89^{th}~term~\hfill }{a_{89}=\cfrac{20}{3}+\cfrac{1}{3}(89-1)}\implies a_{89}=\cfrac{20}{3}+\cfrac{1}{3}(88)\implies a_{89}=\cfrac{20}{3}+\cfrac{88}{3} \\\\\\ a_{89}=\cfrac{20+88}{3}\implies a_{89}=\cfrac{108}{3}\implies a_{89}=36[/tex]
