Question

Ache a soma dos 40 primeiros ternos da P.a (12, 6,..)

Answer 1

$a_1 = 12\\r = -6\\n = 40\\a_n = ?\\S_n = ?\\ \\a_n = a1 + (n-1).r\\a_{40} = 12 + (40-1).(-6)\\a_{40} = 12 - 234\\a_{40} = -222\\ \\ S_n = \frac{(a_1\ +\ a_n).n}{2}\\ \\ S_{40} = \frac{(12\ -\ 222).(-6)}{2}\\ \\ S_{40} = (6 - 111).(-6)\\ \\ S_{40} = -36 + 666\\ \\ S_{40} = 630\\ \\ \\ \\ \boxed{Resposta: \ 630}$