Let the sum of the first n terms of the difference series an be sn, and a3 12, s12 0, and s13 0

1) a1+a12=a6+a7a1+a13=a7*2 can be written as the first term and the tolerance form can be used to prove that s12=(a1+a12)*12 2=(a6+a7)*6s13=(a1+a13)*13 2=a7*13, so a1+2d=12 a6+a7<0, that is, 2a1+11d>0 a7>0, that is, a1+6d<0 is represented by d by the formula A1, that is, a1=12-2d is brought into Eq. respectively: 24+7d>0 12+4d<0 can be solved to obtain -24 70a7<0 to know a6>0, a7<0, and |a6|>|a7|Therefore s1a7>a8>a12, so s6+a7>s7>s8>...

s12, so s6 is the maximum.

1. Solution: a3=12, a1 2d=12, a1=12 2d s12=(a1 a12) 12 2=6(2a1 11d) 0 2a1 11d=24 4d 11d 0 7d 24 d 24 7....and s13=(a1 a13) 13 2=13·a7 0 a7=12 4d 0 d 3....

The value range of 24 7 d 3 d is ( 24 7, 3).

Set the public leakage early surplus difference to d

s12=(a3+a10)*6=(2a3+7d)*6=(24+7d)*6> return to destroy the finch 0

s13=a7*13=(a3+4d)*13=(12+4d)*130 and 12+4d

s12＝（a1+a12）*12/2>0

a1+a12>0,a6+a7>0

s12=(a1+a13)*13/2<0

a1+a13<0,2a7<0,a<>7

It shows that Luyin Xiang S6 is the largest. From item 7 onwards, the SN decreases.

s12=6(a6+a7)>0 a6+a7>0 s13=13*a7-a7

The smallest absolute value is item 7.

s12=12(a1+a12) 2=21 a1+a12=7 2 a3+a4+a9+a10 =(a3+a10)+(a4+a9) =a1+a12)+(a1+a12) =7 pie Li Fang 2+7 2 =7

d=(2sn-24n)/(n*n-5n)

The value range of d is: -4 The value of 3s1 is the largest because the tolerance is less than 0

s20=20

20(a1+a20)/2=10(a1+a20)=10(a1+a1+19d)=20

a3=a1+2d=16

The simultaneous solution yields a1=20 and d=-2

So an=a1+(n-1)d=20-2(n-1)=22-2n, so sn=n(a1+an) 2=n(20+22-2n) 2=n(21-n).

Therefore, s10 = 10 (21-10) = 110

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It is known to be a series of equal differences, and the first term is a1 and the tolerance is d

then a3=a1+2d=16 (1)s20=(a1+a1+19d)*20 2=20 2a1+19d=2 (2).

2)-(1)*2 15d=-30 d=-2 substitute (1) a1=20

s10+(a1+a1+9d)*10/2=5(20+20-18)=110