Question

Insira o setermeios arimetricos entre -7 e 81 .

Answer 1

Inserindo 7 meios aritméticos entre os dois já existentes, a PA ficará com 9 termos.

$PA:~~-7~,~a_1~,~a_2~,~a_3~,~a_4~,~a_5~,~a_6~,~a_7~,~a_8~,~81$

Vamos começar determinando a razão pela equação do termo geral da PA:

$\boxed{a_n~=~a_m+(n-m).r}\\\\\\\\a_9~=~a_1+(9-1).r\\\\\\81~=~-7+8r\\\\\\8r~=~81-(-7)\\\\\\r~=~\frac{88}{8}\\\\\\\boxed{r~=~11}$

Sendo assim, podemos determinar os termos inseridos com a equação do termo geral:

$\rightarrow~a_2~~=~~a_1+(2-1).11~~=~-7+11~~\rightarrow~\boxed{a_2~=~4}\\\\\\\rightarrow~a_3~~=~~a_2+(3-2).11~~=~4+11~~\rightarrow~\boxed{a_3~=~15}\\\\\\\rightarrow~a_4~~=~~a_3+(4-3).11~~=~15+11~~\rightarrow~\boxed{a_4~=~26}\\\\\\\rightarrow~a_5~~=~~a_4+(5-4).11~~=~26+11~~\rightarrow~\boxed{a_5~=~37}\\\\\\\rightarrow~a_6~~=~~a_5+(6-5).11~~=~37+11~~\rightarrow~\boxed{a_6~=~48}\\\\\\\rightarrow~a_7~~=~~a_6+(7-6).11~~=~48+11~~\rightarrow~\boxed{a_7~=~59}$

$\rightarrow~a_8~~=~~a_7+(8-7).11~~=~59+11~~\rightarrow~\boxed{a_8~=~70}$

Ficamos com:

$PA:~~-7~,~4~,~15~,~26~,~37~,~48~,~59~,~70~,~81$

Answer 2

resolução!

an = a1 + ( n - 1 ) r

81 = - 7 + ( 9 - 1 ) r

81 = - 7 + 8r

81 + 7 = 8r

88 = 8r

r = 88 / 8

r = 11

PA = { - 7 , 4 , 15 , 26 , 37 , 48 , 59 , 70 , 81 }