Question

imterpole 8 meios aritméticos entre 12e75?​

Answer 1

r=an-a1/n-1 = 75-12/10-1=63/9= 7

=> 12, 19, 26, 33, 40, 47, 54, 61, 68, 75

Answer 2

Adicionando 8 termos entre os 2 já existentes ficaremos com a seguinte PA de 10 termos:

PA:  12 , a2 , a3 , a4 , a5 , a6 , a7 , a8 , a9 , 75

Vamos determinar a razão desta PA, utilizando a equação do termo geral:

$\boxed{a_n~=~a_1+(n-1).r}\\\\\\\\a_{10}~=~a_1+(10-1).r\\\\\\75~=~12+9r\\\\\\9r~=~75-12\\\\\\r~=~\frac{63}{9}\\\\\\\boxed{r~=~7}$

Com o valor da razão podemos achar os termos inseridos:

$a_2~=~a_ 1+r~=~12+7~~\rightarrow~~\boxed{a_2~=~19}\\\\\\a_3~=~a_ 2+r~=~19+7~~\rightarrow~~\boxed{a_3~=~26}\\\\\\a_4~=~a_ 3+r~=~26+7~~\rightarrow~~\boxed{a_4~=~33}\\\\\\a_5~=~a_ 4+r~=~33+7~~\rightarrow~~\boxed{a_5~=~40}\\\\\\a_6~=~a_ 5+r~=~40+7~~\rightarrow~~\boxed{a_6~=~47}\\\\\\a_7~=~a_ 6+r~=~47+7~~\rightarrow~~\boxed{a_7~=~54}\\\\\\a_8~=~a_ 7+r~=~54+7~~\rightarrow~~\boxed{a_8~=~61}\\\\\\a_9~=~a_ 8+r~=~61+7~~\rightarrow~~\boxed{a_9~=~68}$