等差数列an的前n项和为Sn,若S10=20,S20=240,则当m,n取正数时,代数式(n*am-m*an)/am*an的最大值为?

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等差数列an的前n项和为Sn,若S10=20,S20=240,则当m,n取正数时,代数式(n*am-m*an)/am*an的最大值为?等差数列an的前n项和为Sn,若S10=20,S20=240,则当

等差数列an的前n项和为Sn,若S10=20,S20=240,则当m,n取正数时,代数式(n*am-m*an)/am*an的最大值为?
等差数列an的前n项和为Sn,若S10=20,S20=240,则当m,n取正数时,代数式(n*am-m*an)/am*an的最大值为?

等差数列an的前n项和为Sn,若S10=20,S20=240,则当m,n取正数时,代数式(n*am-m*an)/am*an的最大值为?
a(n) = a + (n-1)d,
s(n) = na + n(n-1)d/2.
20 = s(10) = 10a + 45d,4 = 2a + 9d.2a = 4 - 9d.
240 = s(20) = 20a + 190d.24 = 2a + 19d = 4 - 9d + 19d.20 = 10d.d = 2.
2a = 4 - 9d = - 14.a = -7.
a(n) = -7 + 2(n-1) = 2n - 9.
[na(m) - ma(n)]/[a(m)a(n)] = [n(2m-9) - m(2n-9)]/[(2n-9)(2m-9)]
= [2mn-9n - 2mn+9m]/[(2n-9)(2m-9)]
= 9(m-n)/[(2n-9)(2m-9)]
= (9/2)[2m-9-2n+9]/[(2n-9)(2m-9)]
= (9/2)[1/(2n-9) + 1/(9-2m)]
10)单调递减.1/(2n-9)

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