Is an arithmetic sequence with a tolerance of 0 an arithmetic sequence?

Is an arithmetic sequence with a tolerance of 0 an arithmetic sequence?

Not necessarily. According to the concept of equal ratio sequence, non-zero constant sequence is equal ratio sequence

In the sequence, an-an-1 = constant. Can we prove that the sequence is arithmetic? An / an-1 = constant, can we prove that the sequence is proportional?
RT.

Yes, but it also requires A1 and should be marked (n > = 2)

How to sum up the multiplication of each term of an arithmetic sequence and an arithmetic sequence

The basic methods and skills of sequence summation Wei Fengling key words: sequence summation general term fraction method dislocation subtraction reverse order addition grouping method grouping method combination method sequence is an important part of high school algebra, but also the basis of learning higher mathematics. It plays an important role in college entrance examination and various mathematical competitions

How to find the sum of arithmetic sequence divided by arithmetic sequence?

In short, it is to arrange many groups of equal ratio sequence, sum, and then sum the result

What is the nature of the sum of the first n terms of the arithmetic sequence?

Isochromatic s (2n-1) = (2n-1) an
Sn s2n-sn s3n-s2n into arithmetic sequence
sn-s(n-1)=an

The sum of the three positive numbers in the arithmetic sequence is equal to 15, and the three numbers are added with 2, 5 and 13 respectively to form B3, B4 and b5 in the arithmetic sequence {BN}. (I) find the general formula of the sequence {BN}; (II) the sum of the first n terms of the sequence {BN} is Sn, and prove that the sequence {Sn + 54} is the arithmetic sequence

(1) Let the three positive numbers of the arithmetic sequence be A-D, a, a + D respectively. According to the meaning of the problem, we get A-D + A + D = 15 and a = 5. Therefore, the sequence of {BN} is 7-d, 10, 18 + D. according to the meaning of the problem, we have (7-d) (18 + D) = 100 and d = 2 or D = - 13 (rounding off). Therefore, the third term of {BN} is 5 and the common ratio is 2. From B3 = B1 · 22, that is 5 = 4b1, we get B1 = 54. So {BN} is an arithmetic sequence with the first term of 54 and the common ratio of 2 The formula is the front of BN = 54.2n − 1 (II) sequence {BN} and Sn = 54 (1 − 2n) 1 − & nbsp; 2 = 54.2n − 54, that is, Sn + 54 = 5.2n4, so S1 + 54 = 52, Sn + 1 + 54sn + 54 = 5.2n − 15.2n − 2 = 2, so {Sn + 54} is an equal ratio sequence with 52 as the first term and common ratio of 2

The sum of the three positive numbers in the arithmetic sequence is equal to 15, and these three numbers are added with 2, 5 and 13 respectively to form B2, B4 and b5 in the arithmetic sequence {BN}
(1) Find the general formula of the sequence {BN}; (2) the sum of the first n terms of the sequence {BN} is Sn, and prove that the sequence {Sn + 5 / 4} is an equal ratio sequence (with process)

(1) Let the three positive numbers of arithmetic sequence be A-D, a, a + D respectively
a-d+a+a+d=15,a=5
So the order of {BN} is 7-d, 10,18 + D
Have (7-d) (18 + D) = 100, d = 2 or D = - 13 (rounding off)
So the third term of {BN} is 5 and the common ratio is 2
From B3 = B1 &; 22, that is 5 = 4b1, B1 = 5 / 4
So {BN} is an equal ratio sequence with the first term of 5 / 4 and the common ratio of 2. The general formula is BN = 5 * 2 ^ (n-1) / 4
(2) The first sum of sequence {BN} Sn = (5 / 4) * (1-2 ^ n) / (1-2) = 5 * 2 ^ n / 4-5 / 4
Sn + 5 / 4 = 5 * 2 ^ (n-2), so S1 + 5 / 4 = 5 / 2,
[(Sn+1)+5/4]]/ [Sn+5/4]]=[5*2^(n-1)]/[ 5*2^(n-2)]=2
So {Sn + 5 / 4} is an equal ratio sequence with 5 / 2 as the first term and 2 as the common ratio

The sum of the three positive numbers in the arithmetic sequence is equal to 15. When the second number is added with 10, it becomes the arithmetic sequence again, and finds the original three numbers

This question is not complete. Here are some examples
The sum of the three positive numbers in the arithmetic sequence is 15. When the second number is added with 1 and the third number is added with 10, the arithmetic sequence is formed to find the original three numbers
Let the three positive numbers of the original arithmetic sequence be a, B, C, so a + B + C = 15, and a + C = B
So, B = 5
That is, a + C = 10
And because a, 6, C + 10 are equal ratio sequence, so
a（c+10）=36 ②
It can be solved by simultaneous methods
a. C equals 2,8
So the original three numbers are 2, 5 and 8

Insert two numbers between 2 and 9, so that the first three numbers are equal difference series, and the last three numbers are equal ratio series?

Let these two numbers be x, x + D
2x=2+(x+d)①
(x+d)^2=9x②
X = 4or1 / 4
So d = 2or-7 / 4
So these two numbers are 4 and 6 or 1 / 4 and - 3 / 2

Insert two numbers between 2 and 20, so that the first three numbers are equal proportion series and the last three numbers are equal difference series, then the sum of the two numbers inserted is ()
A. - 4 or 1712b. 4 or 1712c. 4D. 1712

Let this sequence be 2, x, y, 20, then x2 = 2y2y = x + 20, the solution is x = - 4Y = 8 or x = 5Y = 252, so x + y = 4 or 1712, so choose B