an=a1+(n-1)d ,a2=a1+d=-6,a1+a9=0=2a1+8d
d=2,a1=-8
设bn为数列{a3n-1},n=1,b1=a3*1-1=a2=-6,bn=a3n-1=a1+(3n-2)*d=-12+6n
bn为b1=-6,d=6的等差数列,所以数列{a3n-1}的前n项和为=-9n+3n^2
an=a1+(n-1)d ,a2=a1+d=-6,a1+a9=0=2a1+8d
d=2,a1=-8
设bn为数列{a3n-1},n=1,b1=a3*1-1=a2=-6,bn=a3n-1=a1+(3n-2)*d=-12+6n
bn为b1=-6,d=6的等差数列,所以数列{a3n-1}的前n项和为=-9n+3n^2