等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=(a5)^2

等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=(a5)^2
等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=(a5)^2

等差数列{an}是递增数列,前n项和为Sn,且a1,a3,a9成等比数列,S5=(a5)^2
求通项么?
因为
an=a1+(n-1)*d Sn=n*a1+1/2 [n*(n-1)]*d
a3^2=a1*a9 S5=(a5)^2
所以
(1) (a1+2d)^2=a1(a1+8d)
(2) 5*a1+10d=(a1+4d)^2
a1=d=3/5 a1=d=0
又因为an递增,所以d不为0
所以
an=3/5+3/5*(n-1)=3/5*n

d>0
a3=a1+2d
a9=a1+8d
a5=a1+4d
(a3)^2=(a1)^2+4a1d+4d^2
a1*a9=a1*(a1+8d)=(a1)^2+8a1d
已知a1,a3,a9成等比数列
(a3)^2=a1*a9
(a1)^2+4a1d+4d^2=(a1)^2+8a1d
因d>0
故a1=d
(a5...

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d>0
a3=a1+2d
a9=a1+8d
a5=a1+4d
(a3)^2=(a1)^2+4a1d+4d^2
a1*a9=a1*(a1+8d)=(a1)^2+8a1d
已知a1,a3,a9成等比数列
(a3)^2=a1*a9
(a1)^2+4a1d+4d^2=(a1)^2+8a1d
因d>0
故a1=d
(a5)^2=(a1)^2+8a1d+16d^2=25d^2
s5=(a1+a1+4d)*5/2=5a1+10d=15d
已知S5=(a5)^2
15d=25d^2
d=a1=3/5
an=3/5+(n-1)*3/5=3n/5
a5=3*5/5=3
s5=9

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