Question

В арифметической прогрес-
сии a3 = 31 и a7 = 79. Найдите
a11 и S31.

Answer 1

$a_{3} =31; a_{7} =79; \\ a_{3} = a_{1} +2d; \\ a_{7} = a_{1} +6d; \\ -\left \{ {{ a_{1}+2d=31 } \atop {a_{1}+6d=79}} \right. \\ 4d=48 \\ d=12 \\ a_{1} =7 \\ a_{11} =7+120=127 \\ S_{n} = \frac{( a_{1}+ a_{n})n }{2} \\ a_{31} =7+360=367 \\ S_{31} = \frac{(7+367)*31}{2}=187$