Question

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

Answer 1

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