设等比数列前n项和为Sn,公比为q.若S5,S15,S10成等差数列.求证求证2S5,S10,S20-S10成等比数列.

当q=1时，为常数数列；
当q≠1时，根据等差求出q^5为-1/2,然后就可证明了。

设首项是a
则S5=a(q^5-1)/(q-1)
S15=a(q^15-1)/(q-1)
S10=a(q^10-1)/(q-1)
S5,S15,S10成等差数列
所以2a(q^15-1)/(q-1)=a(q^5-1)/(q-1)+a(q^10-1)/(q-1)
2q^15-2=q^5-1+q^10-1
2q^15=q^5+q^10
S20=a(q^20-1)/(q-1)
S20-S10=a(q^20-1)/(q-1)-a(q^10-1)/(q-1)=a(q^20-q^10)/(q-1)
2S5*(S20-S10)/(S10)^2
=2a(q^5-1)/(q-1)*[a(q^20-q^10)/(q-1)]/[a(q^10-1)/(q-1)]^2
=2(q^5-1)(q^20-q^10)/(q^10-1)^2
=2q^10(q^5-1)(q^10-1)/(q^10-1)^2
=2q^10(q^5-1)/(q^10-1)
=(2q^15-2q^10)/(q^10-1)
前面得到2q^15=q^5+q^10
所以(2q^15-2q^10)/(q^10-1)
=(q^5+q^10-2q^10)/(q^10-1)
=(q^5-q^10)/(q^10-1)
=q^5(1-q^5)/(q^5+1)(q^5-1)
=-q^5/(q^5+1)不等于1
题不对吧

题目没问题设首项是a1,公比是qS5+S10=2S15(S中都乘有一项a1/1-q,由于等式两边都有在此略去!)(1-q^5)+(1-q^10)=2(1-q^15)q^5(2q^10-q^5+1)=0∵q≠0∴2q^10-q^5+1=0∴q^5=1或-1/2当q^5=1即q=1时显然2S5=S10=S20-S10成...