Question

Să se găsească primul termen și rația unei progresii geometrice (bn), dacă b5 - b1=15 și b4 - b2=6

Answer 1

[tex]\div \div(b_n)_{n\geq1}\\ \left \{ {{b_5-b_1=15} \atop {b_4-b_2=6}} \right. \Leftrightarrow \left \{ {{b_1\cdot q^4-b_1=15} \atop {b_1\cdot q^3-b_1\cdot q=6}} \right. \Leftrightarrow \left \{ {{b1\cdot(q^2-1)(q^2+1)=15} \atop {b_1\cdot q(q^2-1)=6}} \right. \\ Daca\ inlocuim\ in\ prima\ relatie\ vom\ avea:\\ \frac{6}{q}(q^2+1)=15\\ 6q^2+6=15q\\ 6q^2-15q+6=0\\ 2q^2-5q+2=0\\ \Delta=25-16=9\Rightarrow \sqrt{\Delta}=3\\ q_1=\frac{5+3}{4}=2\\ q_2=\frac{5-3}{4}=\frac{1}{2}\\ i)Daca\ q=2:\\ [/tex]
[tex]b_1\cdot2\cdot(4-1)=6\Rightarrow b_1=1\\ ii)Daca\ q=\frac{1}{2}\\ b_1\cdot\frac{1}{2}(\frac{1}{4}-1)=6\Rightarrow b_1=-16[/tex]

Answer 2

[tex]\it \ b_5-b_1=15 \Rightarrow b_1q^4-b_1=15 \Rightarrow b_1(q^4-1) =15 \Rightarrow \\\;\\ \Rightarrow b_1[(q^2)^2-1^2]=15 \Rightarrow b_1(q^2-1)(q^2+1)=15 \ \ \ \ (1)[/tex]

$\it b_4-b_2 = 6 \Rightarrow b_1q^3-b_1q=6 \Rightarrow b_1q(q^2-1)=6\ \ \ (2)$

Împărțim cele două relații :

[tex]\it \dfrac{b_1(q^2-1)(q^2+1)}{b_q(q^2-1)} = \dfrac{15}{6} \Rightarrow \dfrac{q^2+1}{q}=\dfrac{5}{2} \Rightarrow 2q^2+2 = 5q \Rightarrow \\\;\\ \\\;\\ \Rightarrow 2q^2-5q+2=0 v 2q^2-4q-q+2=0 \Rightarrow2q(q-2)-(q-2) =0 \\\;\\ \Rightarrow (q-2)(2q-1)=0[/tex]