In a series of equal differences, the sum of the first 20 terms is 170,

The first solution is obviously wrong a6+a14=a9+a11≠a1+a20

A1=a should be given with the first n terms and the formula 20a+190d=170 a6+a9+a11+a14=4a+36d=?In this way, the problem is unsolvable, so I think the question is wrong, and the easiest way to do this is to change the first 20 items to the first 19 items, as mentioned in 2, so that you can get the result 34

The answer is wrong, it should be 19 items.

If it's 19, then.

a6+a14=2a10

a9+a11=2a10

So a6+a14+a9+a11 4a10=4(a1+a19) 2=170 19*4

The second one is correct, this is a wrong question, and I will give the following solution.

If it is a series of equal differences or proportional numbers, if it is a multiple-choice question, and there is no line in the selected answer that says none of the above answers are correct, you can completely treat this series as a constant series. That is, the tolerance d=0 and the tolerance q=1. However, there are tips for other questions, and this method is not an absolute method.

And the big question is definitely not appropriate.

But it's not appropriate for you.

For your problem, you can have a very simple way to do, first see a1 + a20 = 17 in this case, we can assume that its common ratio is 1, we can get a1 = -1 a20 = 18 and then calculate the direction formula is an=n-2, and then you can calculate the sum of the above terms is 32, so the original approach is not right, this question can have different answers according to different assumptions, so this question is wrong, for example, if you assume that this question is a constant sequence, you can get the answer is 34.

If a negative number is introduced.

It is possible to start from -1 to 18

If you can't have introduced a negative number, then.

The second one is right.

Because of the difference series.

It's not that it's 1

It can be a decimal.

Will lz should be.

Questions, if it is 170, 20 items.

It's 32, not 34

If it is 19 items.

The answer is 34

...All in half.

If you follow the practice of the first person.

First of all, we can find a1+a20=17--- which is correct.

But a6 + a14 = a1 + a19 --- is not equal to a1 + a20 (a1 + a19 cannot be calculated).

In this problem, there is no way to know the difference between the items, so it is impossible to calculate the sum of a6 + a9 + a11 + a14.

If the first 19 items are replaced and read 170.

a1+a19=2*170 19=340 19--- not 17, so it's only half right

The second pair. Because a(1)+a(4)=a(2)+a(3).

In the same way, a(6)+a(14)=a(9)+a(11)=a(1)+a(19).

The second pair. Because a(1)+a(19)=a(2)+a(18)=a(3)+a(17)=a(4)+a(16)=a(5)+a(15)=a(6)+a(14)=a(7)+a(13)=a(8)+a(12)=a(9)+a(11) and not equal a(1)+a(20).

Therefore, there is an error in this question, so change the sum of the first 19 items to 170.

But the second result seems to be 340 19, not 17

The second pair. The question is that there is a problem.

If you represent the sequence of numbers in letters, let a1=x, and the difference between two places is y, then a1=x, a20=x+19y, then a1+a20=2x+19y 2(a1=a20)=4x+38y

But a6 + a14 + a9 + a11 = 4x + 36y, it is not equal to 2 (a1 = a20).

Explain that the first one is wrong, and the same letter can verify that the second one is right!

According to the first n terms and formulas of the equivalence pairing state series, s10=10a1+45d=155, s20=20a1+95d=610, and the source can be known when the system of equations is solved.

s20-s10=a11+a12+a13+.a20 = 800, so that a11 + a12 + a13 +A20 = B2, the sum of the first 10 terms is B1, their tolerance is 800-100 = 700, it is said to find the sum of the first 50 terms is to find the sum of B1 + B2 + B3 + B4 + B5, the spike section B1 = 100, tolerance = 700, you can find it.

The sum of the first 10 terms of the series is 50, and the sum of the last 10 terms is 60 for the shed, so 10*(a1+an)=50+60=110, so a1+an=11

So sn=n(a1+an) excitator2=11n2

Let the number column be a1,a2,..an-1,annow: a1+a2+a3+a4+an-3+an-2+an-1+an = 40+80 = 120

There is also a1+an = a2+an-1 = a3+an-2 = a4+an-3

SO: A1+AN = 120 4 = 30 while the first n terms and the formula are 210 = (A1+AN)N 2= 30N2 = 15N

n = 14

Let the difference series scatter and have n terms.

S front = s4 = 4 (a1 + a4) 2 = 26

a1+a4=13

After s = 4 (a(n-3) + an) 2 = 110a (n-3) + an = 55

a1+an=a(n-3)+an=13+55=68a1+an=34

sn=n(a1+an)/2=187

Solution, n=11, please click "Pick the hood Huaina Chong Sui as the answer".

According to the equation for summing the difference series: sn=na1+n(n-1)d 2, so there is: 10a1+10(10-2)d 2=2020a1+20(20-1)d 2=60

Solve the system of equations to obtain: a1=, d=

So: s30 = 30a1 + 30 (30-1) d 2 = 30

The few upstairs are not the right steps to solve the problem!

In the equal difference series, the sum of the first 10 terms is 20, and the sum of the first 20 terms is 60, then there is (s20-s10)-s10=(20-10)*(10d)(s30-s20)-s20=(30-20)*(10d)(s20-s10)-s10=(s30-s20)-s20 has s30=3s20-2s10=3*60-2*20=140, then the sum of the first 30 is 140

Let the sum of the first n terms be sn, s10, s20-s10, and s30-s20 also form an equal difference series, and the tolerance is 10 times the tolerance of the original equal difference series.

Because s10=a1+a2+.a10=20s20-s10=a11+a12+..a20=40, so s30-s20=a21+a22+.a30=60s30=120

Tolerance = (67-49) (32-23) = 2

The first term = 49-22*2=5

a8 = 19 a9 = 21 a23 = 49 a24 = 51 ninth to twenty-third terms.

Let the tolerance be d, from the meaning of the question: -20 + (50-1) * d = 120, find d = 20 7, with the formula: (first term + last term) multiply the number of terms divided by 2. That is: (-20+120)*50 2=2500