Найдите сумму первых двенадцати членов арифметической прогрессии (an), если: a6=12; a16=100.

\[a_{6} = 12;\ \ a_{16} = 100:\]

\[a_{6} = a_{1} + 5d \rightarrow a_{1} = a_{6} - 5d;\]

\[a_{16} = a_{1} + 15d \rightarrow a_{1} =\]

\[= a_{16} - 15d.\]

\[12 - 5d = 100 - 15d\]

\[10d = 88\]

\[d = 8,8.\]

\[a_{1} = 12 - 5 \cdot 8,8 = 12 - 44 =\]

\[= - 32.\]

\[S_{12} = \frac{2a_{1} + 11d}{2} \cdot 12 =\]

\[= 6 \cdot ( - 64 + 96,8) = 6 \cdot 32,8 =\]

\[= 196,8.\]

\[Ответ:196,8.\]