Question

En una progresion geometrica de termino positivo
A1=2 A3=6 halla AN,A11,A12



URGENTEEE DOY 35 PUNTOS Y EXPLICARMELO POR FAVORR

Answer 1

Lore,
Vamos paso a paso

En una PG
                  an = a1.q^(n-1)

En el caso en estudio
                                    a3 = a1.q^(3-1)
                                      6 = 2.q^2
                                      3 = q^2
                                      q = √3
                                                        an = 2(√3)^(n-1)

                                                        a11 = 2(√3)^(11-1) = 486

                                                        a12 = 2(√3)^(12-1) = 486√3