Question

设Sn是等差数列{an}的前n项和，已知1/3S3与1/4S4的等比中项为1/5S5，1/3S3与1/4S4的等差中项为1。急啊！~

设Sn是等差数列{an}的前n项和，已知1/3S3与1/4S4的等比中项为1/5S5，1/3S3与1/4S4的等差中项为1，求等差数列{an}的通项an

Answer 1

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与1/4S4的等比中项为1/5S5,
(a1 + 2d)^2 =(a1 + d)×(a1 + 3d/2)
3a1d/2 + 5d^2/2 = 0 ......(1)
1/3S3与1/4S4的等差中项为1,
(a1 + d)+(a1 + 3d/2)=2×1
a1 + 5d/4 = 1......(2)
(1),(2)两式联立，解得：d=0或d=-12/5
当d=0时，代入(2)中，a1=1 ,an=1
当d=-12/5时，代入(2)中，a1=4 ,an=32/5 - 12n/5

Answer 2

S3=3a+3d
S4=4a+6d
S5=5a+10d
由题意得：1/3(3a+3d)*1/4(4a+6d)=[1/5(5a+10d)]^2
1/3(3a+3d)+1/4(4a+6d)=1*2
所以：a=4,d=-12/5
所以：an=4-12/5(n-1)