等差数列的前n项和为48,前2n项和为60,则前3n项和为()A.84 B.72 C.36 D.-24

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等差数列的前n项和为48,前2n项和为60,则前3n项和为()A.84 B.72 C.36 D.-24

等差数列的前n项和为48,前2n项和为60,则前3n项和为()A.84 B.72 C.36 D.-24
等差数列的前n项和为48,前2n项和为60,则前3n项和为()
A.84 B.72 C.36 D.-24

等差数列的前n项和为48,前2n项和为60,则前3n项和为()A.84 B.72 C.36 D.-24
记a_i表示原数列的第i项,
A=a_1+a_2+…+a_n,
B=a_(n+1)+a_(n+2)+…+a_(2n),
C=a_(2n+1)+a_(2n+2)+…+a_(3n),
则B=[a_(n+1)+a_(n+2)+…+a_(2n)]
=[a_1+nd]+{a_2+nd}+…+{a_n+nd]
=A+n^2*d,
同理 C=B+n^2*d
A,B,C组成首项为公差为D=n^2*d的等差数列,
又由题意知 A=48,A+B=60,
所以B=60-48=12,
D=B-A=12-48=-36,
C=B+D=12+(-36)=-24,
因此,数列的前3n项和为A+B+C=60+(-24)=36.
解析:因为是等差数列,所以Sn,S2n-Sn,S3n-S2n也是等差数列

即n+1到2n的和是60-48=12 比前n项的和少12-48=36
2n+1到3n的和是12-36=-24
前3n项的和是60-24=36
选C