已知Sn是等比数列前n项和,S3、S6、S9成等差数列,求证a2 a5 a8为等差数列.三种解法.
证明一
因为S3.S9.S6成等差数列
2S9=S3+S6
2a1(1-q^8)/(1-q)=a1(1-q^2)/(1-q)+a1(1-q^5)/(1-q)
2(1-q^8)=2-q^2-q^5
2q^8=q^2+q^5
2q^7=q+q^4
2a1q^7=a1q+a1q^4
2a8=a2+a5
所以a2.a8.a5成等差数列
证明二：
等比数列{an},各项均不能为0
2S9=S3+S6
(a4+a5+a6+a7+a8+a9)+(a7+a8+a9)=0
(a4+a5+a6)=-2(a7+a8+a9)
(1+q+q^2)=-2(q^3+q^4+q^5)
1=-2q^3
q^3=-1/2
2a8-(a2+a5)=a2*(q^6-1-q^3)=a2*0=0
所以2a8=a2+a5
即a2,a8,a5成等差数列
证明三
证明：由已知设An=A1q^(n-1),q为公比且不为0.
则q不为0也不为1时,Sn=[(q^n-1)/(q-1)]A1；
当q=1时,Sn=nA1；
∵2S9=S3+S6,
∴2[(q^9-1)/(q-1)]A1=[(q^3-1)/(q-1)]A1+[(q^6-1)/(q-1)]A1,
∴2q^9=q3+q^6,∴q3(2q3+1)(q3-1)=0,
∴q3=1或-1/2,∴q=1或(-1/2)^(1/3),
当q=1时,An=A1,则2A8-(A8+A5)=2A1-(A1+A1)=0,得证；
当q=(-1/2)^(1/3)时,An=A1[(-1/2)^(1/3)]^(n-1)；
2A8-(A8+A5)
=2×A1[(-1/2)^(1/3)]^(8-1)
-{[A1[(-1/2)^(1/3)]^(2-1)]+[A1[(-1/2)^(1/3)]^(5-1)]}
=(A1/2)(-1/2)^(1/3)-[A1(-1/2)^(1/3)+(-A1/2)(-1/2)^(1/3)]
=(A1/2)(-1/2)^(1/3)-(A1/2)(-1/2)^(1/3)
=0,得证. 已知Sn是等比数列前n项和,S3、S6、S9成等差数列,求证a2 a5 a8为等差数列.三种解法.
证明一
因为S3.S9.S6成等差数列
2S9=S3+S6
2a1(1-q^8)/(1-q)=a1(1-q^2)/(1-q)+a1(1-q^5)/(1-q)
2(1-q^8)=2-q^2-q^5
2q^8=q^2+q^5
2q^7=q+q^4
2a1q^7=a1q+a1q^4
2a8=a2+a5
所以a2.a8.a5成等差数列
证明二：
等比数列{an},各项均不能为0
2S9=S3+S6
(a4+a5+a6+a7+a8+a9)+(a7+a8+a9)=0
(a4+a5+a6)=-2(a7+a8+a9)
(1+q+q^2)=-2(q^3+q^4+q^5)
1=-2q^3
q^3=-1/2
2a8-(a2+a5)=a2*(q^6-1-q^3)=a2*0=0
所以2a8=a2+a5
即a2,a8,a5成等差数列
证明三
证明：由已知设An=A1q^(n-1),q为公比且不为0.
则q不为0也不为1时,Sn=[(q^n-1)/(q-1)]A1；
当q=1时,Sn=nA1；
∵2S9=S3+S6,
∴2[(q^9-1)/(q-1)]A1=[(q^3-1)/(q-1)]A1+[(q^6-1)/(q-1)]A1,
∴2q^9=q3+q^6,∴q3(2q3+1)(q3-1)=0,
∴q3=1或-1/2,∴q=1或(-1/2)^(1/3),
当q=1时,An=A1,则2A8-(A8+A5)=2A1-(A1+A1)=0,得证；
当q=(-1/2)^(1/3)时,An=A1[(-1/2)^(1/3)]^(n-1)；
2A8-(A8+A5)
=2×A1[(-1/2)^(1/3)]^(8-1)
-{[A1[(-1/2)^(1/3)]^(2-1)]+[A1[(-1/2)^(1/3)]^(5-1)]}
=(A1/2)(-1/2)^(1/3)-[A1(-1/2)^(1/3)+(-A1/2)(-1/2)^(1/3)]
=(A1/2)(-1/2)^(1/3)-(A1/2)(-1/2)^(1/3)
=0,得证.