已知等差数列{An}的前n项,和Sn=n²-9n,第k项满足5<Ak<8,则k为多少?

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已知等差数列{An}的前n项,和Sn=n²-9n,第k项满足5<Ak<8,则k为多少?

已知等差数列{An}的前n项,和Sn=n²-9n,第k项满足5<Ak<8,则k为多少?
已知等差数列{An}的前n项,和Sn=n²-9n,第k项满足5<Ak<8,则k为多少?

已知等差数列{An}的前n项,和Sn=n²-9n,第k项满足5<Ak<8,则k为多少?
Sk = k² - 9k
Sk-1 = (k - 1)² - 9(k - 1)
两式相减得:
Ak = 2k - 10
因为 5 < Ak < 8
所以 5 < 2k - 10 < 8
所以 7.5 < k < 9
所以 k = 8