A column of numbers 2, 5, 8, 11, 14, L7, find the sum of 20 numbers in the middle of the first 100 numbers

The middle 20 numbers are from the 41st to the 60th. The sequence is an arithmetic sequence with the first term of 2 and the variance of 3! The 41st term is 122 and the 60th term is 179. Then calculate the sum of the terms between 122 and 179, which is equal to (122 + 179) x2o / 2 = 3010

Fill in the following brackets with the appropriate number: 6 / 18 = () / () = () / () 4 = () / 2 = () / 6 1 / 6 = () / () = () / () = ()/（

6/18=(3 )/(9 )=(1 )/(3 )
4=（ 8)/2=(24 )/6
1/6=(2 )/（ 12）=（ 3）/（ 18）=（4 ）/（24

Is the neighborhood coprime of Lucas sequence
1,3,4,7,11,…… Are two adjacent terms coprime

It can be simply proved that the term number relationship of Lucas sequence is the same as that of fibulacci sequence, that is, there is a (n + 1) = an + a (n-1). Suppose there are two terms a (n + 1), an is not mutually prime, and there is a common factor D, then it is obvious that a (n + 1) - an = a (n-1), and a (n-1) obviously contains a factor D, that is, there is a (n-1), an is not mutually prime, then recursively, all numbers have a factor D, which is obviously wrong

On sequence
How to find the general formula of 1,2,4,7,11,16

an-a（n-1）=n-1,
a（n-1）- a（n-2）=n-2,
a（n-2）- a（n-3）=n-3,
.
.
.
a2-a1=1；
Add left and right to get
an-a1=1+2+3+.+（n-1）=n（n-1）/2,
So an = n (n-1) / 2 + 1

It's about sequences,
In the sequence an, A1 = 1, an = 2Sn ^ 2 / (2sn-1) (n ≥ 2), then the sum of the first n terms of the sequence is

Because an = SN-S (n-1)
From the condition an = 2Sn ^ 2 / 2S (n-1) (n ≥ 2), we obtain that
Sn-S(n-1)=2(Sn^2)/(2Sn-1) (n≥2),
S (n-1) - sn-2s (n-1) Sn = 0
Divide both sides by s (n-1) Sn
So 1 / sn-1 / S (n-1) = 2
So the sequence {1 / Sn} is an arithmetic sequence, its first term 1 / S1 = 1 / A1 = 1, the tolerance is 2
So: 1 / Sn = 1 + 2 (n-1) = 2N-1
Sn=1/(2n-1)
That is, the sum of the first n terms of the sequence {an} is Sn = 1 / (2n-1)

Some problems about sequence of numbers
If there are only three numbers a, B and C in the sequence, then:
If the square of B = AC, must it be an equal ratio sequence?
If 2B = a + C, is this sequence an arithmetic sequence?
That is to say, if ABC is not zero, it is an equal ratio sequence?

If there are only three numbers a, B and C in the sequence, then:
If the square of B = AC, must it be an equal ratio sequence? No, it is not an equal ratio sequence when it is equal to 0
If 2B = a + C, is this sequence an arithmetic sequence? Yes
That is to say, if ABC is not zero, it is an equal ratio sequence?
Yes, 1, 1, 1, or 2, 2 is an equal ratio sequence with a common ratio of 1

For an arithmetic sequence with even terms, the sum of odd terms is 24, and the sum of even terms is 30. If the last term is 21 / 2 larger than the first term,
Find the first term, tolerance and number of terms of this sequence

a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2.
a(2n)=a+(2n-1)d=a+d+(n-1)(2d),
a(2n-1)=a+(2n-2)d=a+(n-1)(2d).
30=b(n)=a(2)+a(4)+...+a(2n)=n(a+d)+n(n-1)d.
24=c(n)=a(1)+a(3)+...+a(2n-1)=na+n(n-1)d.
21/2=a(2n)-a(1)=(2n-1)d.
6=30-24=nd.
d=2nd-21/2=2*6-21/2=3/2.
n=6/d=4.2n=8.
24=na+n(n-1)d=4a+4*3*3/2=4a+18,
6=4a,a=3/2.
First item a = 3 / 2, tolerance d = 3 / 2, number of items 2n = 8

It is known that there is an arithmetic sequence whose terms are even. The sum of odd and even terms is 24 and 30 respectively. If the difference between the last term and the first term is 10.5, find this
The first term, tolerance and number of terms of a sequence

Let the first term, tolerance and number of terms of the arithmetic sequence be A1, D, n.an = a1 + (n-1) DN, and S (odd) = n * A1 / 2 + (0 + n-2) * n * D / 4 = Na1 / 2 + n (n-2) d / 4 = 24s (even) = n * A1 / 2 + (1 + n-2) * n * D / 4 = Na1 + (n-1) nd / 4 = 30an-a1 = (n-1) d = 10.5

For an arithmetic sequence with even terms, the sum of odd and even terms is 24 and 30 respectively. If the last term is 21 / 2 larger than the first term, the number of terms in the sequence is?

Let this sequence have 2n terms, 30-24 = nd
nd=6
a2n-a1=(2n-1)d=21/2
2nd-d=21/2
d=3/2
24=n(a1+a2n-1)/2
30=n(a2+a2n)/2=n(a1+a2n-1+2d)/2=n(a1+a2n-1)/2+nd=24+n(3/2)
3n/2=6
3n=12
n=4

If the sum of odd and even terms of an even arithmetic sequence is 24 and 30 respectively, and the last one is 21 / 2 larger than the first one, then the last one is 24

Let 2n terms be added
Then nd = 30-24 = 6
[a1+(2n-1)d]-a1=21/2
2nd-d=21/2
So d = 2nd-21 / 2 = 3 / 2
n=6/d=4
a1+a3+a5+a7=24
So 4A1 + 12D = 24
a1=3/2
So A8 = a1 + 21 / 2 = 12

A column of numbers 2, 5, 8, 11, 14, L7, find the sum of 20 numbers in the middle of the first 100 numbers