Question

Interpole 8 meios aritméticos entre 2 e 38.

Answer 1

$P.A.\ (2,\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ 38)$

Pelo enunciado é deduzido que $a_1=2$ e $a_{10}=38$. Com esses valores podemos calcular a razão da P.A.:

$a_{10}=a_{1}+9r\\\\38=2+9r\\\\36=9r\\\\r=\dfrac{36}{9}\\\\r=4$

Desta forma, por conseguinte podemos interpolar os oito meios aritméticos:

$a_2=a_1+r\longrightarrow{a}_2=2+4\longrightarrow{a}_2=6\\\\a_{3}=a_{2}+r\longrightarrow{a}_{3}=6+4\longrightarrow{a}_{3}=10\\\\a_{4}=a_{3}+r\longrightarrow{a}_{4}=10+4\longrightarrow{a}_{4}=14\\\\a_{5}=a_{4}+r\longrightarrow{a}_{5}=14+4\longrightarrow{a}_{5}=18\\\\a_{6}=a_{5}+r\longrightarrow{a}_{6}=18+4\longrightarrow{a}_{6}=22\\\\a_{7}=a_{6}+r\longrightarrow{a}_{7}=22+4\longrightarrow{a}_{7}=26\\\\a_{8}=a_{7}+r\longrightarrow{a}_{8}=26+4\longrightarrow{a}_{8}=30\\\\a_{9}=a_{8}+r\longrightarrow{a}_{9}=30+4\longrightarrow{a}_{9}=34$

Portanto, tem-se a seguinte P.A.:

$P.A.\ (2,\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ \underline{\ \ \ \ },\ 38)\\\\\boxed{P.A.\ (2,\ 6,\ 10,\ 14,\ 18,\ 22,\ 26,\ 30,\ 34)}$