Avem suma termenilor unei progresii aritmetice cu rația 3.
[tex]\it a_n=a_1+(n-1)r=1+(n-1)\cdot3=1+3n-3=3n-2=91|_{+2} \Rightarrow \\ \\ \Rightarrow 3n=93|_{:3} \Rightarrow n=31\ (num\breve arul\ termenilor\ sumei)\\ \\ \\ S_{31}=\dfrac{(a_1+a_{31})\cdot 31}{2}=\dfrac{(1+91)\cdot31}{2}=\dfrac{92\cdot31}{2}=46\cdot31=1426[/tex]
