9.设数列{an}和{bn}的前
本题应该是等差数列(——否则答案不确定)!
法一:
a1 + a15 = a1 + a1 + (15-1)d = 2a1 + 14d = 2(a1 + 7d) = 2a8
a8 / b8
= (2*a8) / (2*b8)
= (a1 + a15) / (b1 + b15)
= [15 * (a1 + a15)/2] / [15 * (b1 + b15)/2]
= S15 / T15
= (7*15 + 2) / (15 + 3)
= 107/18
法二:
因为等差数列的前n项和 Sn 可以写成下面的形式:
Sn = n*a1 + n(n-1)d/2 = (d/2)n² + ...全部
本题应该是等差数列(——否则答案不确定)!
法一:
a1 + a15 = a1 + a1 + (15-1)d = 2a1 + 14d = 2(a1 + 7d) = 2a8
a8 / b8
= (2*a8) / (2*b8)
= (a1 + a15) / (b1 + b15)
= [15 * (a1 + a15)/2] / [15 * (b1 + b15)/2]
= S15 / T15
= (7*15 + 2) / (15 + 3)
= 107/18
法二:
因为等差数列的前n项和 Sn 可以写成下面的形式:
Sn = n*a1 + n(n-1)d/2 = (d/2)n² + (a1 - d/2)n = A*n² + B*n
所以由 Sn/Tn = (7n+2)/(n+3)
可设 Sn = kn(7n+2) , Tn = kn(n+3) (k是常数)
于是 a8 = S8 - S7 = k*8(7*8+2) - k*7(7*7+2) = 。
。。k
b8 = T8 - T7 = k*8(8+3) - k*7(7+3) = 。。。。。。。k
所以 a8 / b8 = 。。。k / 。。。。。。k = 107/18
。收起