[tex]\it3)\ \ \ r=\dfrac{a_9-a_3}{6}=\dfrac{4-9}{6}=-\dfrac{5}{6}\\ \\ \\ a_n=a_3+(n-3)r=9+(n-3)\cdot(-\dfrac{5}{6}) =9-\dfrac{5n-15}{6} =\dfrac{69-5n}{6}\\ \\ \\ Sau:\\ \\ \\ a_n=a_9+(n-9)r=4+(n-9)\cdot(-\dfrac{5}{6}) =4-\dfrac{5n-45}{6}=\dfrac{69-5n}{6}[/tex]
