某等差數列的第2項和第5項分別是75和51。設該數列的首n項之和為Sn。 a) 求該數列的最小正數項。?

等差數列第2項和第5項分別是75和51。首n項之和為Sn。求數列的最小正數項。


a2 = a + d = 75

a5 = a + 4d = 51


相減: 3d = -24


=> d = -8

=> a = 75 - d = 75 + 8 = 83


an = 83 - 8(n-1) > 0

=> 83 + 8 > 8n

=> 8n < 92

=> n < 92/8 = 11.5

=> a11 = 83 - 8*10 = 3 = Answer