Найдите сумму первых десяти членов арифметической прогрессии (an), если: a4=16; a12=88.

Ответ:

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

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

\[a_{4} = 16;\ \ a_{12} = 88:\]

\[a_{4} = a_{1} + 3d \rightarrow a_{1} = a_{4} - 3d;\]

\[a_{12} = a_{1} + 11d \rightarrow a_{1} =\]

\[= a_{12} - 11d.\]

\[a_{4} - 3d = a_{12} - 11d\]

\[16 - 3d = 88 - 11d\]

\[8d = 72\]

\[d = 9.\]

\[a_{1} = a_{4} - 3d = 16 - 3 \cdot 9 =\]

\[= 16 - 27 = - 11.\]

\[S_{10} = \left( 2 \cdot ( - 11) + 9 \cdot 9 \right) \cdot 5 =\]

\[= ( - 22 + 81) \cdot 5 = 59 \cdot 5 =\]

\[= 295.\]