Question

Va rog din tot sufletul problema 31*, am nevoie URGENTTT ​

Answer 1

$\it Fie\ \ r\ -\ ra\c{\it t}ia\ \ progresiei\ \ aritmetice\\ \\ a_1+a_2+a_3=12 \Rightarrow a_2-r+a_2+a_2+r=12 \Rightarrow 3a_2=12 \Rightarrow a_2=4\\ \\ a_1a_2a_3=28 \Rightarrow a_1\cdot4\cdot a_3=28|_{:4} \Rightarrow a_1\cdot a_3=7 \Rightarrow (a_2-r)(a_2+r)=7 \Rightarrow \\ \\ \Rightarrow (4-r)(4+r)=7 \Rightarrow 16-r^2=7 \Rightarrow r^2=16-7 \Rightarrow r^2=9 \Rightarrow r=\pm3$

Avem două progresii aritmetice:

$\it I)\ \ a_1=a_2-r=4-(-3)=4+3=7,\ \ \ r=-3\\ \\ II)\ \ a_1=a_2-r=4-3=1,\ \ r=3$

$\it a_7-a_5=a_1+6r-(a_1+4r)=a_1+6r-a_1-4r=2r=\pm6$