Question

Calcule a soma dos doze primeiros termos da PA ( -3, -1, 1, 3,...).

Answer 1

$\text{A\ soma\ dos\ n\ primeiros\ de\ uma\ PA\ é:} \\ \\ \boxed{S_{n} =\frac{( a_{1} +a_{n}) n}{2}} \\ \\ r=a_{2} -a_{1} =-1-( -3) \Longrightarrow r=2 \\ \\ a_{12} =a_{1} +11r=-3+11\cdotp 2\Longrightarrow a_{12} =19 \\ \\ \therefore S_{12} =\frac{( a_{1} +a_{12}) \cdotp 12}{2} =( -3+19) \cdotp 6 \\ \\ \Longrightarrow \boxed{\boxed{S_{12} =96}}$