Question

determinati primul termen si ratia progresiei geometrice daca$\left \{ {{b4-b2=18} \atop {b5-b3=36}} \right.$

Answer 1

Se exprima si aici toti termenii in functie de primul si ratia progresiei geometrice, cu formula:  $b_{n}= b_{1}* q^{n-1}$

Answer 2

$\it b_4-b_2 = 18 \Longrightarrow b_2q^2-b_2=18 \Longrightarrow b_2(q^2-1)=18\ \ \ \ (1)$

$\it b_5-b_3 = 36 \Longrightarrow b_3q^2-b_3=36 \Longrightarrow b_3(q^2-1)=36\ \ \ \ (2)$

$\it (1),\ (2) \Longrightarrow \dfrac{b_3}{b_2} =\dfrac{36}{18} \Longrightarrow \dfrac{b_3}{b_2} = 2 \Longrightarrow \dfrac{b_2q}{b_2}=2 \Longrightarrow q = 2 \ \ \ \ (3)$

$\it (1), (2) \Longrightarrow b_2(2^2-1) =18 \Longrightarrow b_2\cdot3=18 \Longrightarrow b_2 =6 \Longrightarrow b_1q=6$

$\it \Longrightarrow b_1\cdot2=6 \Longrightarrow b_1=3$