公差不为零的等差数列{an}的前几项和为Sn,若a4是a3与a7的等比中项,S8=32,求首项a1等于多少?公差d(2)

由题可知(a4)^2=a3*a7,及等差数列公式an=a1+(n-1)d,Sn=n(a1+an)/2.
(a1+3d)^2=(a1+2d)(a1+6d),d(2a1+3d)=0,又d≠0,所以2a1+3d=0,
a1=-3d/2.又S8=8(a1+a8)/2=32,8(a1+a1+7d)/2=32,(2a1+7d)/2=4,
结合a1=-3d/2,得a1=-3,d=2.
S10=10(a1+a10)/2=5(a1+a1+9d)=5(2a1+9d)=5[2*(-3)+9*2]=60.