等差数列An,前N项和为Sn,求证S4,S8-S4,S12-S8成等差数列 若等差改成等比,是否仍有此性质?

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等差数列An,前N项和为Sn,求证S4,S8-S4,S12-S8成等差数列若等差改成等比,是否仍有此性质?等差数列An,前N项和为Sn,求证S4,S8-S4,S12-S8成等差数列若等差改成等比,是否

等差数列An,前N项和为Sn,求证S4,S8-S4,S12-S8成等差数列 若等差改成等比,是否仍有此性质?
等差数列An,前N项和为Sn,求证S4,S8-S4,S12-S8成等差数列 若等差改成等比,是否仍有此性质?

等差数列An,前N项和为Sn,求证S4,S8-S4,S12-S8成等差数列 若等差改成等比,是否仍有此性质?
2(S8-S4)=2(8a1+28d-4a1-6d)=8a1+44d
S4+S12-S8=4a1+6d+12a1+66d-8a1-28d=8a1+44d
2(S8-S4)=S4+S12
S4,S8-S4,S12-S8成等差数列.
若S4、S8-S4、S12-S8成等比数列,则
(S8-S4)²=S4×(S12-S8)
(4a1+22d)²=(4a1+6d)(12a1+66d)
整理,得
(2a1+11d)^2=(2a1+3d)(6a1+33d)
4a1^2+20a1d-11d^2=0
(2a1+11d)(2a1-d)=0
a1=-11d/2或a1=d/2
即只有当a1=-11d/2或a1=d/2,且a1、d都不等于0时,才满足等比数列成立.

感觉他回答的好像不对，因为等比数列的前Sn=an(1-q*n）/1-q或Sn=a1-anq/1-q,而他带的是等差数列公式，是不对的

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