若等差数列中a1+a2+a3+a4+a5+a6=10,S12=40,求S24...
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若等差数列中a1+a2+a3+a4+a5+a6=10,S12=40,求S24...
若等差数列中a1+a2+a3+a4+a5+a6=10,S12=40,求S24...
若等差数列中a1+a2+a3+a4+a5+a6=10,S12=40,求S24...
等差数列的性质
S6,S12-S6,S18-S12,S24-S18成等差数列
S6=10
S12-S6=30
∴ S18-S12=50
S24-S18=70
∴ S24=S6+S12-S6+ S18-S12+S24-S18=10+30+50+70=160
S6=10;S12=40
S12-S6=a7+...+a12=(a1+6d)+...(a6+6d)=s6+6dx6=s6+36d 所以 d=5/9
s24-s12=a13+...a24=(a1+12d)+...(a12+12d)=s12+12dx12=40+144x5/9=120
s24=160
a6=a1+5d
a12=a1+11d
a24=a1+23d
a1+a2+a3+a4+a5+a6=10
1/2×6(a1+a6)=10
6a1+15d=10 (1)
S12=40
1/2×12(a1+11d)=40
12...
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a6=a1+5d
a12=a1+11d
a24=a1+23d
a1+a2+a3+a4+a5+a6=10
1/2×6(a1+a6)=10
6a1+15d=10 (1)
S12=40
1/2×12(a1+11d)=40
12a1+66d=40 (2)
(2)-(1)×2得:
36d=20 d=5/9
a1=5/18
S24=1/2×24(a1+a24)
=12(a1+a1+23d)
=24a1+276d
=24×5/18+276×5/9
=1600
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[1]6a1+15d=10 (1) 12a1+66d=40 (2)
[2]6a1+15d=10 (1)
6a1+33d=20 (2)
[3]a1=5/18 d=5/9
[4]S24=24a1+12*23d=
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