Sn=na1 + [n(n-1)d]/2
1/3S3=1/3{3a1 + [3×(3-1)]d/2} = 1/3(3a1 + 3d)= a1 + d
1/4S4=1/4{4a1 + [4×(4-1)]d/2} = 1/4(4a1 + 6d)= a1 + 3d/2
1/5S5=1/5{5a1 + [5×(5-1)]d/2} = 1/5(5a1 + 10d)= a1 + 2d
1/3S3與1/4S4的等比中項(xiàng)為1/5S5,
(a1 + 2d)^2 =(a1 + d)×(a1 + 3d/2)
3a1d/2 + 5d^2/2 = 0 .(1)
1/3S3與1/4S4的等差中項(xiàng)為1,
(a1 + d)+(a1 + 3d/2)=2×1
a1 + 5d/4 = 1.(2)
(1),(2)兩式聯(lián)立,解得：d=0或d=-12/5
當(dāng)d=0時(shí),代入(2)中,a1=1 ,an=1
當(dāng)d=-12/5時(shí),代入(2)中,a1=4 ,an=32/5 - 12n/5