等差数列an中,a1=13,S3=S11,那么此数列前几项和最大?最大值是?
热心网友
Sn = A1*n + n*(n-1)*d/2A1 = 13, S3 = S11 ==== 13*3 +3*(3-1)*d/2 = 13*11 + 11(11-1)*d/2=== d = -2Sn = A1*n + n*(n-1)*d/2 = 13*n + n*(n-1)*(-2)/2 = -n^2 + 14*n= -(n-7)^2 + 49<= 49 ( n = 7)此数列前7项和最大, 最大值是49
热心网友
∵S3=3[2a1+(3-1)d]/2,S11=11[2a1+(11-1)d]/2,(这里的d是公差)，S3=S11,∴3[2a1+(3-1)d]/2=11[2a1+(11-1)d]/2.解得d=-(2/13)a1=-2.∴Sn=n[2a1+(n-1)d]/2=n[26-2(n-1)]/2=-n^2+14n=-(n-7)^2+49.∴当n=7时，Sn最大，且Sn最大=49.即此数列前7项的和最大，且这个最大值为49.