Найдите сумму десяти первых членов арифметической прогрессии (an), если: a1=6, a13=42.

\[a_{1} = 6;\ \ \ \ \ \ a_{13} = 42:\]

\[S_{10} = \frac{2a_{1} + 9d}{2} \cdot 10 =\]

\[= \left( 2a_{1} + 9d \right) \cdot 5 = 10a_{1} + 45d.\]

\[a_{13} = a_{1} + 12d\]

\[12d = a_{13} - a_{1}\]

\[12d = 42 - 6\]

\[12d = 36\]

\[d = 3.\]

\[S_{10} = 10 \cdot 6 + 45 \cdot 3 =\]

\[= 60 + 135 = 195.\]

\[Ответ:195.\ \]