Question

等差数列{an}的前n项为Sn,且S3=6,a1=4,则公差d=多少?要详解

Answer 1

S3=a1+a2+a3
= a1+a1+d + a1+2d
= 3a1 + 3d
=6
又如汪因为a1=4
所谨好以12+3d=6
解得渣晌仔d=-2

Answer 2

设公差d
s3=3*a1+3d=6
3*4+3d=6
12+3d=6
3d=-6
d=-2

Answer 3

S3=A1+A2+A3
=A1+A1+d+A1+2d
=3A1+3d

=> 6=3A1+3d
=>腊穗键族誉3d=-6
=>轮巧d=-2

Answer 4

等差数列前N项和公式S=(A1+An)N/州镇雀册早2
等旅首差数列公式求和公式 Sn=n(a1+an)/2 或Sn=na1+n(n-1)d/2
S3=3*(a1+a3）/2 =6,a3=0
a(n)=a1+(n-1)d
a3=a1+2d,d=-2

Answer 5

S3=a1+a2+a3=a1+(a1+d)+(a1+2d)
6=3a1+3d d=-2