Question

1- qual a razão da P.A. (4,-2,-8,...)?

2- encontre o termo geral da P.A. (4,-2,-8,...)?

3- qual o 20º termo da P.A. (5,7,9,11,13,...)?

4- qual a posição do termo 42 na P.A (4,6,8,...)?

Answer 1

1.

$(a_1;a_2;a_3;\dots)=(4;-2;-8;\dots)\\\\ \boxed{a_n=a_k+(n-k)r}\ \therefore\ a_2=a_1+r\ \therefore\\\\ r=a_2-a_1=-2-4\ \therefore\ \boxed{r=-6}$

2.

$(4,-2;-8;\dots);\ r=-6\\\\ \boxed{a_n=a_k+(n-k)r}\ \therefore\ a_n=a_1+(n-1)(-6)\ \therefore\\\\ a_n=4-6n+6\ \therefore\ \boxed{a_n=10-6n}$

3.

$(a_1;a_2;a_3;a_4;a_5;\dots)=(5;7;9;11;13;\dots)\\\\ \boxed{r=\dfrac{a_n-a_k}{n-k}}\ \therefore\ r=\dfrac{a_2-a_1}{2-1}=7-5\ \therefore\ \boxed{r=2}\\\\ \boxed{a_n=a_k+(n-k)r}\ \therefore\ a_n=a_1+(n-1)(2)\ \therefore\\\\ a_n=5+2n-2\ \therefore\ \boxed{a_n=3+2n}\\\\ a_{20}=3+2(20)\ \therefore\ \boxed{a_{20}=43}$

4.

$(a_1;a_2;a_3;\dots;a_{42};\dots)=(4;6;8;\dots)\\\\ \boxed{r=\dfrac{a_n-a_k}{n-k}}\ \therefore\ r=\dfrac{a_2-a_1}{2-1}=6-4\ \therefore\ \boxed{r=2}\\\\ \boxed{a_n=a_k+(n-k)r}\ \therefore\ a_n=a_1+(n-1)(2)\ \therefore\\\\ a_n=4+2n-2\ \therefore\ \boxed{a_n=2(n+1)}\\\\ a_n=42\ \therefore\ 42=2(n+1)\ \therefore\ 21=n+1\ \therefore\ \boxed{n=20}$