Question

insira quatro meios aritméticos entre 3 e 243. calcule a soma.

Answer 1

(3,...........243)
a1,a2,a3,a4,a5,a6

a6=a1+5r
243=3+5r
243-3=5r

5r=240
r=240/5
r=48

an=a1+(n-1)r
an=3+(n-1)48
an=3+48n-48
an=48n-45

a2=48x2-45= 51
a3=48x3-45=99
a4=48x4-45=147
a5=48x5-45=195

PA(3,51,99,147,195,243)