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

题目

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

答案

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) ...