Question

determine a razão da P.A.(-5,5,-5,5,-5,5) e a soma dos seis primeiros termos​

Answer 1

$PG=(a_1;a_2;\dots;a_6;\dots)=(-5;5;-5;5;-5;5;\dots)\\\\ \text{Termo geral}\ \to\ \boxed{a_n=a_kq^{n-k}}\\\\ a_2=a_1q^{2-1}\ \therefore\ a_2=a_1q\ \therefore\ 5=-5q\ \therefore\ \boxed{q=-1}\\\\ \text{Soma dos n primeiros termos}\ \to\ \boxed{S_n=a_1\dfrac{q^n-1}{q-1}}\\\\ S_6=-5\dfrac{(-1)^6-1}{-1-1}=-5\dfrac{1-1}{-2}\ \therefore\ \boxed{S_6=0}$