在递减的等差数列an中,前n项和为Sn,且满足a19*a21=1260,a14+a16+a18+a22+a24+a26=
在递减的等差数列an中,前n项和为Sn,且满足a19*a21=1260,a14+a16+a18+a22+a24+a26=216,在满足Sn>an的所有n中取最大整数N,求Sn
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an= a1+(n-1)d
a14+a16+a18+a22+a24+a26=216
6a1+114d =216
a1+19d =36 (1)
a19.a21=1260
(a1+18d)(a1+20d)=1260
(36-d)(36+d) =1260
d^2=36
d=-6
a1 = 150
an = 150-6(n-1) =156-6n
Sn = (153-3n)n
Sn > an
(153-3n)n > 156-6n
3n^2- 159n+156
a14+a16+a18+a22+a24+a26=216
6a1+114d =216
a1+19d =36 (1)
a19.a21=1260
(a1+18d)(a1+20d)=1260
(36-d)(36+d) =1260
d^2=36
d=-6
a1 = 150
an = 150-6(n-1) =156-6n
Sn = (153-3n)n
Sn > an
(153-3n)n > 156-6n
3n^2- 159n+156
1年前
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