\[a_{1};\ \ a_{1} + d;\ \ a_{1} + 2d;\ \ldots.\]
\[3a_{1};\ \ 3a_{1} + 3d;\ \ 3a_{1} + 6d;\ \ldots.\]
\[3a_{1} + 3d = \frac{3a_{1} + 3a_{1} + 6d}{2} =\]
\[= \frac{6a_{1} + 6d}{2} = 3a_{1} + 3d \Longrightarrow да,\ \]
\[будет.\]