Question

Знайдіть суму двадцяти перших членів арифметичної прогрессії

Answer 1

$\displaystyle\bf a_{19} = 60 \\ a_{29} = 95 \\ a_{n} = a_{1} + (n - 1)d \\ \displaystyle\bf\\\left \{ {{a_{19} = a_{1} + 18d} \atop {a _{29} =a_{1} + 28d}} \right. \\ \displaystyle\bf\\\left \{ {{a_{1} + 18d = 60 \: \: | \times ( - 1) } \atop {a _{1} + 28d = 95 }} \right. \\ \displaystyle\bf\\ + \left \{ {{ - a_{1} - 18d = - 60} \atop { a _{1} + 28d = 95}} \right. \\ \\ 28d - 18d = 95 - 60 \\ 10d = 35 \\ d = 35 \div 10 \\ d = 3.5 \\ \\ a_{1} + 18 \times 3.5 = 60 \\ a_{1} + 63 = 60 \\ a_{1} = 60 - 63 \\ a_{1} = - 3 \\ \\ a_{12} = a_{1} + 11d = - 3 + 11 \times 3.5 = \\ = - 3 + 38.5 = 35.5 \\ \\ S_{n} = \frac{a_{1} + a_{n}}{2} \times n \\ S_{12} = \frac{a_{1} + a_{12}}{2} \times 12 = 6(a_{1} + a_{12}) = \\ = 6( - 3 + 35.5) = 6 \times 32.5 = 195$