It is known that {an} is an arithmetic sequence, the sum of the first n terms is Sn, A4 = 2, S5 = 20. Let TN = | A1 | + | A2 | + +|An |, find TN

It is known that {an} is an arithmetic sequence, the sum of the first n terms is Sn, A4 = 2, S5 = 20. Let TN = | A1 | + | A2 | + +|An |, find TN

If the tolerance is D, then: A1 = 2 - 3D, A2 = 2 - 2D, A3 = 2 - D, A4 = 2, A5 = 2 + D ∵ S5 = 20 ∵ 10 - 5D = 20 d = - 2 ∵ an = 10 - 2n, hope to help you
Unit conversion problem ruler, inch, Zhang
I want to know 1m =? Ruler
1m =? Inch
1m = 3 feet
1 Zhang = 10 feet
1 foot = 10 inches
1m = 3 feet = 30 inches = 0.3 feet
It is a unit of length in ancient China
1m = 3 feet, 1m = 30 inches, 1m = 0.3 Zhang
The result of 5 2's addition, subtraction, multiplication and division is equal to 8. Each of the four symbols can only be used once
Add subtract multiply divide each can only be used once
(2/2+2)*2+2=8
2×2×2×2÷2＝8
2(2+2)×2÷2=8
Let the sum of the first n terms of the arithmetic sequence {an} be SN. If S4 ≥ 10 and S5 ≤ 15, then the maximum value of A4 is______ .
The sum of the first n terms of {an} is Sn, and S4 ≥ 10, S5 ≤ 15, S4 = 4A1 + 4 × 32D ≥ 10s5 = 5A1 + 5 × 42d ≤ 15, that is, 2A1 + 3D ≥ 5A1 + 2D ≤ 3, A4 = a1 + 3D ≥ 5 − 3D2 + 3D = 5 + 3d2a4 = a1 + 3D = (a1 + 2D) + D ≤ 3 + D, 5 + 3D2 ≤ A4 ≤ 3 + D, 5 + 3D ≤ 6 + 2D, D ≤ 1, A4 ≤ 3 + D
Unit to inch?
How big is an inch?
One inch = 25.4 mm
1 inch = 0.0254 M
=
5 * (6 + 1) (square of 6 + 1) (cubic power of 6 + 1)... (1006th power of 6 + 1) + 1
5 * (6 + 1) (square of 6 + 1) (cubic power of 6 + 1)... (1006th power of 6 + 1) + 1
=(6-1)(6+1)(6²+1)...(6^1006+1)+1
=(6²-1)(6²+1)...(6^1006+1)+1
=.
=(6^1006-1)*(6^1006+1)+1
=6^2012-1+1
=6 ^ 2012 (the power of 6 in 2012)
Repeated use of square difference formula
This problem uses the group sum method of sequence
5×[（6﹢1）﹙6²+1﹚﹙6³+1﹚+。。。。 +(1006 power of 6 + 1)] + 1
=5×[（6﹢6²﹢6³﹢。。。 + 6 to the 1006th power) + 1 × 1006] + 1
=5 × [6 (1006 power of 1-6) / (1-6)] + 1006 × 5 + 1
= - 6 (1006 power of 1-6) + 5030 + 1
=The 1007 power of 6 + 5... Expansion
This problem uses the group sum method of sequence
5×[（6﹢1）﹙6²+1﹚﹙6³+1﹚+。。。。 +(1006 power of 6 + 1)] + 1
=5×[（6﹢6²﹢6³﹢。。。 + 6 to the 1006th power) + 1 × 1006] + 1
=5 × [6 (1006 power of 1-6) / (1-6)] + 1006 × 5 + 1
= - 6 (1006 power of 1-6) + 5030 + 1
=6 to the power of 1007 + 5025
The sum of the first n terms of the arithmetic sequence {an} is known, the tolerance D ≠ 0, and S3 + S5 = 50, A1, A4, A13 are equal ratio sequence
The sum of the first n terms of the arithmetic sequence {an} is known, the tolerance D ≠ 0, and S3 + S5 = 50, A1, A4, A13 is equal ratio sequence. Find the general term of {an}. Sn = A1 * n + [n (n-1) D] / 2 S3 = 3A1 + [3 (3-1) D] / 2 = 3A1 + 3ds5 = 5A1 + [5 (5-1) D] / 2 = 5A1 + 10d, because S3 + S5 = 50, so 3A1 + 3D + 5A1 + 10d = 50, namely: 8A1
s3+s5=3a1+3d+5a1+10d=8a1+13d=50
a1a13=a4^2
a1(a1+12d)=(a1+3d)^2
2a1d=3d^2
a1=3d/2
8a1+13d=50
D=2
a1=3
an=a1+(n-1)d=3+2(n-1)=2n+1
How many centimeters is 17 inches in units?
Buy a LCD display that is 17; I don't know how many centimeters 17 is
17 in = 43.18 cm
The monitor should measure the diagonal length
Find 40 positive and negative number addition, subtraction, multiplication and division problems in the first day of junior high school! Be sure to do: 1. There are addition, subtraction, multiplication and division in a formula, at least 3 must do: 1
Seek 40 first day positive and negative number addition, subtraction, multiplication and division calculation problem!
Must do: 1. A formula in addition, subtraction, multiplication and division, at least 3
As far as possible: 1
1. - 38) + 52 + 118 + (- 62) = 2, (- 32) + 68 + (- 29) + (- 68) = 3, (- 21) + 251 + 21 + (- 151) = 4, 12 + 35 + (- 23) + 0 = 5, (- 6) + 8 + (- 4) + 12 = 6, 3 and 1 / 4 + (- 2 and 3 / 5) + 5 and 3 / 4 + (- 8 and 2 / 5) = 7, 9 + (-)
It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, the tolerance D ≠ 0, and S3 + S5 = 50, A1, A4, A13, which is proportional to the sequence (2)
(2) If we take the second term, the fourth term and the eighth term from the sequence {an} The second ^ n term forms a new sequence {BN} according to the original order to find the first n terms of {BN} and TN
1. The sum formula of arithmetic sequence: SN = Na1 + n (n-1) d / 2, S3 + S5 = (3 + 5) a1 + (3 + 10) d = 8A1 + 13D = 50;
In addition, we can get the following sequence: a 1 + 2 D = a 1 + 2 D, i.e., a 1 + 3 D = a 1 + 2 D = a 1
2、b1=a2=2*2+1,b2=a4=2*4+1,b3=a8=2*8+1,...,bn=2*2^n+1.Tn=ba+b2+b3+...+bn=(2*2+1)+(2*4+1)+(2*8+1)+...+(2*2^n+1)=2*(2+4+8+...+2^n)+n=2^(n+2)+n-4
d=2 a4=9 an=2n+1
bn=2^n+1） +1
Tn=n+4(2^n-1)