设首首项为a
前十项和S10=110 则(a+a+9d)×10=110
a1,a2,a4为等比数列 则(a+d)^2=a(a+3d)
a=11,d=0.an=11
或a=d=2 an=2n公差d的值和数列{2^*an}，的前n项和Tna=11，d=0.an=11，因为d不等于0，舍去 或a=d=2 an=2n2^*an=2^2n=4^n Tn=n×4^n2^*an是2的an次方吧不是可是是我打错了 应该是 {2^an}，是2的n次方在×an的前n项和Tna=d=2 an=2nan×2^n=2n×2^n=n×2^n+1Tn=1×2^2+2×2^3+···+n×2^n+1=(2^2+2^3+···+2^n+1)+(2^3+2^4+···+2^n+1)+···+(2^n+2^n+1)+(2^n+1)=(2^n+2 -2^2)+(2^n+2 -2^3)+···+(2^n+2 -2^n)+(2^n+2 -2^n+1)=n×2^n+2 -(2^2+2^3+···+2^n+1)=n×2^n+2-(2^n+2 -2^2)=(n-1)×2^(n+2) +4