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。?
回答 (1)
2015-09-15 12:15 pm
✔ 最佳答案
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
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