设公差为d,则
an=a1+(n-1)d=dn+a1-d
Sn=na1+n(n-1)d/2
∴Sn/a²n=[na1+n(n-1)d/2]/(dn+a1-d)²
∴lim（n-->∞）Sn/（an）²
=lim（n-->∞）[na1+n(n-1)d/2]/(dn+a1-d)²
=lim（n-->∞）[na1+n(n-1)d/2]/(d²n²+2d(a1-d)n+(a1-d)²]
=lim（n-->∞）[a1/n+(1-1/n)d/2]/(d²+2d(a1-d)/n+(a1-d)²/n²]
=(d/2)/d²=1/(2d)
∵lim（n接近无限大）Sn/（an）²=1/6
∴1/(2d)=1/6 ∴d=3