设等差数列 首项为 a1,等差为 d; a1 = a1 a2 = a1 + d a3 = a1 + 2d . . an = a1 + (n-1)d 等差数列的求和公式是:n(a1+an)/2 代入求和公式,得出两元联立方程: (1) 50(a1+ a50)/2=200 => 50(a1+ a1+49d)/2=200 => 50(2a1+49d)/2=200 (2) 50(a50+ a100)/2=2700 => 50(a1+49d+a1+99d)/2=2700 => 50(2a1+148d)/2=2700 解方程,得: a1 = - 2054/99 d = 100/99 故首项是 -2054/99 |