Question

interpole 8 meios aritméticos entre -40 e 5

Answer 1

Sejam - 40, __, __, __, __, __, __, __, __ e 5.

Temos que:
$\begin{cases}a_1=-40\\a_n=5\\n=10\\r=\end{cases}$

 Então,

$a_n=a_1+(n-1)r\\\\5=-40+(10-1)r\\\\5+40=9r\\\\9r=45\\\\\boxed{r=5}$


 Obtemos o segundo termo somando a razão ao primeiro termo, veja,

$a_2=a_1+r\\\\a_2=-40+5\\\\\boxed{a_2=-35}$

 E assim,...

$a_3=a_2+r\\\\a_3=-35+5\\\\\boxed{a_3=-30}$

$a_4=a_3+r\\\\a_4=-30+5\\\\\boxed{a_4=-25}$

$...$


 Logo,

$\boxed{\boxed{\boxed{\left\{-40,-35,-30,-25,-20,-15,-10,-5,0,5\right\}}}}$