Question

quantos termos tem a pg ( -1/120,1/60,-1/30,...,64/15)

Answer 1

$Dados\to \left\{\begin{array}{ccc}\boxed{a_n = a_1*q^{n-1}}\\\\a_1 = -\frac{1}{120} \\\\a_n = \frac{64}{15} \\\\q = -2\\\\n=\ ?\end{array}\right$

$Resolu\c{c}\~ao\to \left\{\begin{array}{ccc} \frac{64}{15} = -\frac{1}{120}*(-2)^{n-1} \\\\(-2)^{n-1} = \frac{ \frac{64}{15} }{ -\frac{1}{120} }\\\\(-2)^{n-1} = \frac{64}{15}*-120\\\\(-2)^{n-1} = -512\\\\(-2)^{n-1} = (-2)^{9}\\\\n-1 = 9\\\\\boxed{\boxed{n = 10}} \end{array}\right$

Espero ter ajudado. :))