Question

Найдите an и d арифметической прогрессии у которой a1=-3, n=11, Sn=1452

Answer 1

$S_{11}=\frac{a_{1}+a_{11}}{2}*11\\\\1452=\frac{-3+a_{11} }{2}*11\\\\(a_{11}-3)*11=2904\\\\a_{11}-3=264\\\\a_{11}=267\\\\a_{11} =a_{1}+10d\\\\10d=a_{11}-a_{1}=267+3=270\\\\d=27$

Answer 2

$S _{11} = \frac{a_{1} +a_{11}}{2} \times 11 \\ 1452 = \frac{ - 3 +a_{11} }{2} \times 11 \\ 2904 = - 33 + 11a_{11} \\ 11a_{11} = 2904 + 33 \\ 11a_{11} = 2937 \\ a_{11} = 267$

$a _{11} = a _{1} + 10d \\ 267 = - 3 + 10d \\ 10d = 270 \\ d = 27$

Ответ:

$a _{11} = 267 \\ d = 27$