If the first 3 terms of the difference series an are known to be 2,6,10, then a1 a3 a2n 1

Trust me, that's right.

Method 1: When there are 2n terms in the equal difference series, the sum of the even terms - the sum of the odd terms = nd (i.e. n * tolerance) and: the sum of the even terms + the sum of the odd terms = the sum of the number series (i.e. the sum of the first 2n terms) So: the sum of the series = 2 * the sum of the odd terms + nd

So: the sum of odd terms (a1+a3+..a2n-1) = 1 2 [the sum of the sequences (a1 + a2 + a3 + ...)a2n)+nd]=4n*n-2n

Method 2: Set b1=a1,b2=a3,..bn=a2n-1bn} is a series of equal differences with 2 as the first term and 8 as the tolerance, then.

bn=b1+(n-1)*8=8n-6

So: (a1+a3+..a2n-1)=b1+b2+b3+..bn)=1/2(b1+bn)n=4n*n-2n

What's the mess upstairs?

an] is a series of equal differences, and the tolerance is 4, so a1, a3, a5....a2n-1 is the first series of equal differences with an term of 2 and a tolerance of 8, set to [bn].

then bn=8n-6

a1+a3+..a2n-1=b1+b2+b3...bn=s(bn)=4n^2-2n

2, 6, 10 illustrate the difference of 4

The number of items is (a2n-1+2) 4

a1+a3+..a2n-1=(a1+a2n-1)(a2n-1+2)/(2*4)

a2n-1=2+4*(2n-1-1);

a1+a2+a3+..a2n-1=[2+2+4(2n-1-1)][2+4*(2n-1-1)+2] (2*4) simplification.

2n+8(n-1)(n-1) squared will not be absolutely accurate, but I don't believe you can bring a few trees to try.

a2=a1+d=3 a1=13

So d = -10

a8=a1+(8-1)d=13+7*(-10)=-57s8=n(a1+a8) Gao Zheng Sancongqing 2=8*(13-57) Qi2=-176

Summary. a2=5,d=3

In the series of equal differences, a2 = 5 and of = 3, then a10 = a2 = 5 and d = 3

a1=2a10=a1+(n-1)✖️d

a10=2+(10-1)✖️3=29

an=a1+(n-1)✖️d

Solution: Let the tolerance be d.

a25-a10=15d=-22-23=-45d=-3

a1 = a10-9d = 23-9 (-3) = 50 an=a1 + (n-1) d = 50 + (-3) (n-1) = 53-3n order an>0 53-3n>0

3n<53

n<53 3, and n is a positive integer, n 17, that is, the first 17 terms of the number series "0, starting from the 18th term, each term is < 0.

n 17, sn=|a1|+|a2|+.an|=a1+a2+..an

na1+n(n-1)d/2

50n-3n(n-1)/2

n(103-3n)/2

At n 18, sn=|a1|+|a2|+.an|=a1+a2+..a17 -a18-a19-..

an=-(a1+a2+..an) +2(a1+a2+..a17)=-n(103-3n)/2 +2×17×(103-3×17)/2=n(3n-103)/2 +884

a10=a1+9d=23

a25=a1+24d=-22

Solution: a1=50

d=-3a1+(k-1)d<0

i.e., 53-3k<0

i.e. k>17

The 18th term in this sequence starts to be less than 0.

Let the formula for the general term be: an=a1+(n-1)d

a10=a1+9d=30

a25=a1+24d=-15

a25-a10=15d=-45

d--3a1=57

an=57-(n-1)(-3)=60-3n, so the non-negative terms are added to get the maximum, so that an=0, 60-3n=0, n=20 i.e., a20=0, that is, the sum of the first 19 or 20 terms is the maximum.

n 20, tn=n*a1-3n(n-1) 2n>20, starting from term 21, can be regarded as another series of equal differences, bk=|a(k+20)|=3k

k 1) tn = 340 + 3n (k-1) 2 (note n = k + 20).

tn=340+3(n-20)(n-21)/2.

Analysis: According to the equal difference series, a2=4, a6=12, find the first term a1 and the tolerance d, you can find the sum of the first 10 terms.

Answer: Solution: In the equal difference series, a2=4, a6=12a6-a2=4d=12-4=8

d = 2a1 = a2 - d = 4-2 = 2, a10 = a2 + 8d = 4 + 16 = 20 The sum of the first 10 terms is s10 = 5 (a1 + a10) = 5 field hall (2 + 20) = 110

So the answer is: 110

Comments: This question spine excavation examines the problem of knowing three and finding two in the equal difference series, and the cherry spine kernel can generally consider using the basic quantity method, that is, using the known conditions to find the first term and tolerance of the equal difference series, and the problem can be solved It is a simple problem.

Equal difference series {an

Middle, a24, a6

Code file A6A2

4d=12-4=8d=2a1

a2d=4-2=2，a10

a28d=4+16=20

The sum of the first 10 items is Xunmo or Mu Wu S10

5（a1a10

So the answer is: 110

Because the dead branch is an equal difference series, so a1+a2+a3 a1+a1+d+a1+2d 6, the solution is: d 1So use the formula to get an n

Defeated Silver knows A1 and D, using the formula SN Old Banquet.

If the difference is equal, a3 + a5 = 2 a4 = 12

a4=6s6=(a1+a6)*6/2=30

a1+a6=10

Equal difference, so a1+a6=a3+a4=10

So a3=4

Because a2+a4=2a3

So a2=2

Because the sum of the first 6 terms is 30, the difference is 30 5 6 or -6 case 1, a3 + a5 = 12

a3-a5=12

a3＝6a2=6-6=0

The column of difference is -6 0 6 12 18