Question

sete meios aritméticos entre 28 e 140

Answer 1

a1,a2,a3,a4,a5,a6,a7,a8,a9

a9=a1+8r
140=28+8r
r=14

a2=a1+r =28+14=32
a3=a1+2r=28+2x14=56
a4=a1+3r=28+3x14=70
a5=a1+4r=28+4x14=84
a6=a1+5r=28+5x14=98
a7=a1+6r=28+6x14=112
a8=a1+7r=28+7x14=126

PA(28,32,56,70,84,98,112,126