解答

S10=10(2a1+9d)/2=265
2a1+9d=53...(1)
S20=20(2a1+19d)/2=1030
2a1+19d=103...(2)
(2)-(1)
-> 10d=50
d=5 a1=4
S30=30(2a1+29d)/2=2295

我就是我

Sn=n(2a1+(n-1)d)/2
S2n=2n(2a1+(2n-1)d)/2=2n(2a1+(n-1)d)/2 + n²d=2Sn+n²d=100
S3n=3n(2a1+(3n-1)d)/2=3n(2a1+(n-1)d)/2 + 2n²=3Sn+2n²d=150

2Sn+n²d=100...(1)
3Sn+2n²d=150...(2)
(1)*2-(2)
-> Sn=50

我就是我

更正
S3n=3n(2a1+(3n-1)d)/2=3n(2a1+(n-1)d)/2 + 2n²d=3Sn+2n²d=150

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