Question

calcule a soma dos doze primeiros termos da P.G (-3,6,...)​

Answer 1

$\large\boxed{\boxed{{S_{n}=\frac{a_{1}.(q^n)-1}{q-1}}}}}}$

Onde:

a1 = —3

n = 12

q = a2/a1 = 6/-3 = —2

an=a12 =????

Substuindo:

$S_{n}=\frac{-3.(-2^{12}-1)}{-2-1}$

$S_{n}=\frac{-3.(4096-1)}{-3}$

$S_{n}=\frac{-3.4095}{-3}$

$S_{n}=\frac{-2285}{-3}$

$\large\boxed{\boxed{{S_{n}=4095}}}}}$

::::::::Espero ter ajudado:::::::::