daca intr-o progresie geometrica
b2=6 , b5=48 calculati b1, q ,b7,sn
urgeeeeeeeeent


Răspuns :

bn = b1*q^(n-1)

b2 = 6 = b1*q
b5 = 48 = b1*q^4
b5/b2 = 48/6 = b1*q^4/b1*q = q^3
q^3 = 8
q = 2

b2 = 6 = b1*q = b1*2
b1 = 6:2
b1 = 3

b7 = b1*q^6 = 3*2^6 = 3*64 = 192

Sn = b1*(q^n-1)/(q-1) = 3*(2^n-1)/ (2-1) = 3*(2^n-1)