Question

Rezolvati exercitiul dat . Progresie geometrică. ( punctele a și b ) , vă rog urgent ​

Answer 1

$b_{n} = b_{1} * q^{n-1}\\ b_{3} = b_{1} * q^{2}\\ b_{4} = b_{1} * q^{3}\\ b_{2} = b_{1} * q\\ \\ \left \{ {{b_{3} - b_{1} = 12 } \atop {b_{4} - b_{2} = 36 }} \right.\\ \\\left \{ {{b_{1} * q^{2} - b_{1} = 12 } \atop {b_{1} * q^{3} - b_{1} * q = 36}} \right. \\\\b_{1}*q^{2} - b_{1} = 12 = > b_{1} = \frac{12}{q^2-1}$

Inlocuim in a doua ecuatie:

$\frac{12q^2}{q^2-1} - \frac{12q}{q^2-1} = 36\\ \frac{12q^3-12q}{q^2-1} = 36\\ \frac{12q(q^2-1)}{q^2-1} = 36\\ 12q = 36\\q = 3$

Inlocuim q in ecuatia lui b1 si apoi in primele ecuatii

$b_1 = \frac{12}{9-1} = \frac{3}{2} = > \\b_3 = \frac{3}{2} * 9 = \frac{27}{2}\\ b_4 = \frac{3}{2} * 27 = \frac{81}{2}\\ b2 = \frac{3}{2} * 3 = \frac{9}{2}$

Analog se rezolva si pct b