Question

в арифмитической прогрессии сумма первого и четвертого членов равна 16, а сумма второго и шестого членов прогрессии, равна 22. найти сумму первых двенадцати членов арифметической прогрессии

Answer 1

$\displaystyle\bf a_{n} = a_{1} + (n - 1)d\\ \\\left \{ {{a_{1} + a_{4} = 16} \atop {a_{2} + a_{6} = 22}} \right. \\ \displaystyle\bf\\\left \{ {{a_{1} + a_{1} + 3d = 16} \atop {a_{1} + d + a_{1} + 5d = 22 }} \right. \\ \displaystyle\bf\\\left \{ {{2a_{1} + 3d = 16 \: \: | \times ( - 1) } \atop {2a_{1} + 6d = 22 }} \right. \\\displaystyle\bf\\ + \left \{ {{ - 2a_{1} - 3d = - 16} \atop {2a_{1 } + 6d = 22}} \right. \\ \\ 6d - 3d = 22 - 16 \\ 3d = 6 \\ d = 6 \div 3 \\ d = 2 \\ \\ 2a_{1} + 3 \times 2 = 16 \\ 2a_{1} + 6 = 16 \\ 2a_{1} = 16 - 6 \\ 2a_{1} = 10 \\ a_{1} = 10 \div 2 \\ a_{1} = 5 \\ \\ a_{12} = a_{1} + 11d = 5 + 11 \times 2 = 5 + 22 = 27 \\ S_{n} = \frac{a_{1} + a_{n}}{2} n \\ S_{12} = \frac{a_{1} + a_{12}}{2} \times 12 = 6(a_{1} + a_{12}) = \\ = 6 \times (5 + 27) = 6 \times 32 = 192$