等差数列{a}的前n项和为Sn,若S12=84,S20=460,求S28

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等差数列{a}的前n项和为Sn,若S12=84,S20=460,求S28
等差数列{a}的前n项和为Sn,若S12=84,S20=460,求S28

等差数列{a}的前n项和为Sn,若S12=84,S20=460,求S28
设an=a1+(n-1)d
sn=n[2a1+(n-1)d]/2
s12=12(a1+11d/2)=84,a1+11d/2=7
s20=20(a1+19d/2)=460,a1+19d/2=23
d=(23-7)/4=4
a1=7-44/2=-15
s28=28[(-15)+27*4/2]=1092

S12+S28=S40
S12+S28=2S20
S28=2S20-S12
=460*2-84
=836

由 S12=12a1+12*11d/2=84
S20=20a1+20*19d/2=460
列方程组解之得：a1=-15,d=4.
故S28=-15*28+28*27*4/2=1092.

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