Question

已知Sn是等比数列｛an｝的前N项和,S5,S15,S10成等差数列,求证A2,A12,A7成等差数列

Answer 1

S5,S15,S10成等差数列,

2S15=S5+S10

2a1(1-q^15)/(1-q)=a1(1-q^5)/(1-q)+a1(1-q^10)/(1-q)

2-2q^15=1-q^5+1-q^10

2q^15=q^5+q^10

除以q^4

2q^11=q+q^6

2a1q^11=a1q+a1q^6

2a12=a2+a7

所以a2,a12,a7成等差数列