Question

Daca intr-o progresie geometrica (bn) n>1 se cunosc:

b2 = 6, b5 = 48 calculati b1, q, b7, Sn

Answer 1

Stiind ca bn=bx*q(n-x) => b5=b2*q^3 => b2*q^3=48 stiind ca b2=6 => 6*q^3=48 =>q^3=8 => q=2
Stiind: q=2 si b2=6  b1=b2/q=6/2=3
b7=b5*r^2=48*2^2=48*4=192
Sn=b1*(q^n-1/q-1)=3*(2^n-1)/(2-1)=3*(2^n-1)

Answer 2

b₁ ≠ 0        si  q≠ 0 
b₂ = b₁· q                          6= b₁· q
b₅ = b₁· q⁴                        48 = b₁· q⁴          , sunt diferite de zero , le impartim 
48 :6 = b₁· q⁴ : b₁· q = ( b₁ :b₁)· ( q⁴ :q)             ; 8 = q³            ; q³ = 2³ 
ratia  q = 2 
b₁ = 6 : q = 6 : 2 = 3 
b₁ = 3 
b₇ = b₁· q⁶ = 3 ( 2⁶ ) 
Sn= b₁ + b₂ + ... + bn = b₁· [ q ( la n) - 1 ] / ( q -1) 
         = 3 · [  2 ( la n ) - 1 ]