Question

interpolar 6 meios aritméticos entre 7e 42 de modo que a1=7 e a8 = 42

Answer 1

an = a1 + ( n -1) . r  =  an

a1 = 7 + ( 1 -1) .5  =  7
a2 = 7 + ( 2 -1) .5  = 12
a3 = 7 + ( 3 -1) .5  = 17
a4 = 7 + ( 4 -1) .5  = 22
a5 = 7 + ( 5 -1) .5  =  27
a6 = 7 + ( 6 -1) .5  =  32
a7 = 7 + ( 7 -1) .5 =  37
a8 = 7 + ( 8 -1) .5 =  42


PA =  ( 7, 12, 17, 22, 27, 32, 37, 42)

Answer 2

$Calcular~~a~~raz\tilde{a}o \\ \\ a_n=a_8=42 \\ a_1=7 \\ n=6+2=8~~(6~~que~~interpola~+2~das~laterais) \\ r=? \\ \\ a_n=a_1+(n-1)r \\ \\ 42=7+(8-1)r \\ 42=7+7r \\ 42-7=7r \\ 35=7r \\ r=35\div7 \\ \\ r=5$


$a_1=7 \\ a_2=a_1+r=7+5=12 \\ a_3=a_1+2r=7+10=17 \\ a_4=a_1+3r=7+15=22 \\ a_5=a_1+4r=7+20=27 \\ a_6=a_1+5r=7+25=32 \\ a_7=a_1+6r=7+30=37 \\ a_8=a_1+7r=7+35=42 \\ \\ \\ P.A.=(7,\underline{12},\underline{17},\underline{22},\underline{27},\underline{32},\underline{37},42)$