等差数列｛an｝中,a1,a3,a9顺次组成等比数列,公差d≠0,且前10项和S10=110,求数列｛an｝通项公式及前n项和公式
问题描述：
等差数列｛an｝中,a1,a3,a9顺次组成等比数列,公差d≠0,且前10项和S10=110,求数列｛an｝通项公式及前n项和公式
答
a1,a3,a9成等比数列，于是：
a3^2=a1*a9
(a1+2d)^2=a1(a1+8d),化简后得
4d(d-a1)=0
所以a1=d
Sn=n/2*(a1+an)
s10=5(a1+a10)=5(a1+a1+9d)=5*11d=110
于是d=2，a1=2
所以an=a1+(n-1)d=2n
sn=n/2*[a1+a1+(n-1)d]=n(n+1)
答
a(n) = a + (n-1)d.s(n) = na + n(n-1)d/2.[a(3)]^2 = [a+2d]^2 = a(1)a(9) = a*[a+8d], a^2 + 4ad + 4d^2 = a^2 + 8ad,0 = 4d^2 - 4ad = 4d(d-a).a=d.110 = s(10) = 10a + 45d = 55d, d = 2 = a.a(n) = 2 + 2(n-1) ...