Question

Stiind ca ca sirul (bn) n>1 este o progresie geometrica de ratie q:
a) Daca b1 = 1, bn = -128, n = 8 sa se afle: q si Sn
b) Daca q = $\frac{1}{3}$, bn = $\frac{1}{81}$, n = 4 sa se afle b1 si Sn

Answer 1

a.  b₁  = 1 
     b₈ = b₁· q⁷           ; q⁷ = - 128 = ( - 2)⁷  
atunci  q = - 2 
S₈ = b₁· [q⁸ - 1 ]  /  [ q -1 ] = 1· [ ( -2)⁸ - 1 ] / [ - 2 - 1 ] = ( 2⁸ -1 )  / ( -3) 
          = ( 1 - 2⁸ ) / 3 
b.       q =1 /3 
            b₄ = b₁·q³            ;        b₁· ( 1 /3)³  = 1 /81        ; b₁· 1 /27 = 1 / 81 
b₁ = 1 /3 
S₄ = b₁· [ q⁴ -1 ] / [ q -1 ] = 1 /3· [ ( 1/ 3)⁴ - 1 ]  / [ 1 /3 - 1 ] = 
       = [ ( 1 /3)⁴ - 1 ] / ( - 2) = [ 1 - ( 1 /3)⁴ ] / 2 = [ 81  - 1 ] / 81· 2 
         = 80 / 81 ·2 = 40 /81

Answer 2

$\displaystyle a).b_1=1,b_n=-128,n=8,q=?,S_n=? \\b_8=-128 \Rightarrow 1 \cdot q^{8-1}=-128 \Rightarrow 1 \cdot q^7=-128\Rightarrow 1 \cdot q^7=(-2)^7 \Rightarrow \\ \Rightarrow q=-2 \\ S_8=1 \cdot \frac{(-2)^8-1}{-2-1} \Rightarrow S_8= \frac{256-1}{-3} \Rightarrow S_8=\frac{255}{-3}\Rightarrow S_8=- 85$
$\displaystyle b).q= \frac{1}{3} ,b_n= \frac{1}{81} , n=4,b_1=?,S_n=? \\ b_4= \frac{1}{81} \Rightarrow b_1 \cdot \left( \frac{1}{3} \right)^{4-1} = \frac{1}{81} \Rightarrow b_1 \cdot \left(\frac{1}{3} \right)^3= \frac{1}{81} \Rightarrow b_1 \cdot \frac{1}{27} = \frac{1}{81}\Rightarrow \\ \Rightarrow b_1= \frac{1}{3 }$
$\displaystyle S_4= \frac{1}{3} \cdot \frac{\left( \frac{1}{3} \right)^4-1}{ \frac{1}{3} -1} \Rightarrow S_4= \frac{1}{3} \cdot \frac{ \frac{1}{81}-1 }{ \frac{1-3}{3} } \Rightarrow S_4= \frac{1}{3} \cdot \frac{ \frac{1-81}{81} }{- \frac{2}{3} } \Rightarrow \\ \Rightarrow S_4= \frac{1}{3} \cdot \frac{- \frac{80}{81} }{- \frac{2}{3}} \Rightarrow S_4= \frac{1}{3} \cdot \frac{80}{81} \cdot \frac{3}{2} \Rightarrow S_4= \frac{240}{486}\Rightarrow S_4= \frac{40}{81}$