若在等比数列{an}中,前n项和为Sn,若a2,a4,a3成等差数列,判断S2,S4,S3是否成等差数列,并给出证明.

(a1)*q+(a1)*q^2=2*(a1)*q^3 (q≠0)
q+q^2=2*q^3 (q≠0)
1+q=2*q^2
2q^2-q-1=0
q=1,或q=-1/2
q=1时，a1=a2=a3=a4
所以S2=2a1,S4=4a1,S3=3a1,所以q=1时S2,S4,S3不成等差数列
q=-1/2时，S2=[a1*[1-(-1/2)^2]]/[1-(-1/2)]=3*a1/5
S4=[a1*[1-(-1/2)^4]]/[1-(-1/2)]=15*a1/20=3*a1/4
S3=[a1*[1-(-1/2)^3]]/[1-(-1/2)]=7*a1/10
2*S4=3*a1/2, S2+S3=13*a1/10
所以q=-1/2时，S2,S4,S3不成等差数列
综上，S2,S4,S3不成等差数列

若a2,a4,a3成等差数列则2a4=a2+a3所以2a2*q^2=a2+a2*q即2q^2-q-1=0所以q=-1/2或q=1(1)若q=-1/2则S2=a1+a2=a1-a1/2=a1/2S3=S2+a3=a1/2+a1*(-1/2)^2=3a1/4S4=S3+a4=3a1/4+a1*(-1/2)^3=5a1/8所以2S4=S2+S3(2)若q=1则S2=...