Question

Daca intr-o progresie geometrica $(b_{n})_{n\ \textgreater \ 1}$ se cunosc:
a) $b_{1}$ = 2, $b_{2}$ = $\frac{2}{3}$ calculati q, $b_{14}$, $S_{5}$.

Answer 1

in prog. geom  q = b₂ / b₁  = b₅ /b₄ = ........ 
q = b₂ /b₁ = 1/ 3  
b₁₄ = b₁ q¹³ = 2 ( 2 / 3)¹³ = [2¹ 2¹³ ] / 3¹³ = 2¹⁴ /3¹³ 
S₅ = b₁ [ (q⁵ - 1) / ( q -1) ] = 2 [ ( 1 /3 )⁵ - 1 ] / [ 1 /3 - 1 ] =
         = 2 [ (1 /3)⁵ - 1 ] / ( - 2 /3) 
          = - 3 [ (1 /3)⁵  - 1 ] 
           = 3  -  3 / 3⁵
             = 3 - 1 /3⁴  = (3⁵ - 1) / 3⁴ = ( 243 - 1) / 81 = 242  / 81 

Answer 2

$\displaystyle b_1=2,b_2=\frac{2}{3},q=?,b_{14}=?,S_5=?\\q= \frac{\frac{2}{3}}{2} \Rightarrow q= \frac{2}{3} :2 \Rightarrow q= \frac{2}{3} \cdot \frac{1}{2} \Rightarrow q=\frac{2}{6} \Rightarrow q= \frac{1}{3} \\ b_{14}=2 \cdot \left( \frac{1}{3} \right)^{14-1} =2 \cdot \left( \frac{1}{3} \right)^{13}= \frac{2^{14}}{3^{13}} \\ S_5=2 \cdot \frac{\left( \frac{1}{3} \right)^5-1}{ \frac{1}{3}-1}=2\cdot \frac{ \frac{1}{243}-1}{-\frac{2}{3} }=2\cdot\frac{- \frac{242}{243} }{- \frac{2}{3} }= \frac{121}{81}$