Question

设数列An是等差数列,Sn为前n项和。若a1=4,S3/3和S4/4的等比中项为S5/5,求An

Answer 1

∵S3=3a1 + 3d
S4=4a1 + 6d
S5=5a1 + 10d
∴S3/3=4+d , S4/4=4+(3/2)d , S5/5=4+2d
∵S3/3和S4/4的等比中项为S5/5
∴(4+d)[4+(3/2)d] = (4+2d)^2
(5/2)d^2 + 6d=0
∴d=0 或 d=-12/5
∴An=a1+(n-1)d = 4+0=4 ,(n∈N)
或An=a1+(n-1)d = 4 + (n-1)(-12/5) = 32/5 - (12/5)n ,(n∈N)

Answer 2

S3÷3＝4＋d
S4÷4＝4＋3/2d
S5÷5＝4＋2d
(4＋d)(4＋3/2d)＝(4＋2d)²
5d²＋12d＝0
得d＝0 或d＝－12/5
∴An{an＝4，n为正整数}
或An{an＝(8－3n)4/5，n为正整数}

Answer 3

等差中公差和第一项是重点。突破它就好了，还有就是公式