为什么S4=a1+a1+d+a1+2d+a1+3d?

为什么S4=a1+a1+d+a1+2d+a1+3d? 3a1+3d=-3,a1(a1+d)(a1+2d)=8 a1+4d=10 a1+(a1+d)+(a1+2d)=3如何解这个方程组求详细步骤 已知(a1+2d)(a1+6d)=-12与(a1+3d)+(a1+5d)=-4为方程组,求a1与d? {10=a1+2d,a1+（a1+d）+（a1+2d）=3,怎么解得a1=-2,d=3 若等差数列中,S12=8S4,d≠0,则a1/d=? 等差数列an中,若s12=8s4,且d不等于0,则a1/d等于? 设|A|是三阶矩阵,A=(a1,a2,a3)则|A|=?A.|a1-a2,a2-a3,a3-a1| B.|a1-a2,a2-a3,a3-a1|C.|a1+2a2,a3,a1+a2| D.|a1-a3,a2+a3,a1+a2| 等差数列｛an}的公差d不等于0,且a1,a2,a3成等比数列,求（a1+a3+a9)/(a2+a4+a10)的值.为什么(a1+2d)^2=a1*(a1+8d) 在等差数列{An}中,若 S12=8*S4,则A1/d=? a1+2d=-6和a1+4d=-2方程组 解得a1和d是多少 解方程组,1：a1+2d=-3 ,2：a1+4d+a1+9d=30 最好加以说明 计算数列求 5a1+5(5-1)d/2(等差公式)=4(a1+2d)+6与(a1+2d)^2=a1*(a1+8d)帮忙详细解一下a1和d 等差数列a1^2=a11^2,d =VLOOKUP(A1,D:E,2, 已知{an}是等差数列,a1+a2=4,a7+a8=28,则该数列前10项和S10等于多少?设公差为d则a1+a2=a1+a1+d=2a1+d=4 (1)a7+a8=a1+6d+a1+7d=2a1+13d=28 (2)(2)-(1) 12d=24 d=2代入(1) a1=1所以S10=(2a1+9d)*10/2=(2+9*2)*5=100为什么S10=(2a1+9d)*10/2 等差数列公式Sn = a1+a2+...+anSn = a1+(a1+d)+(a1+d+d)+...+[a1+(n-1)d]Sn = a1*n+[1+2+...+(n-1)]*dSn = a1*n+n*(n-1)/2*dSn = a1*n+n*(n-1)*d/2Sn = (a1+an)*n/2an = a1+(n-1)*dSn = [2a1+(n-1)*d]*n/2Sn = a1*n+n*(n-1)*d/2 在等差数列{an}中,若S12=8S4,且公差d不等于0,则d分之a1等于多少?