Find the number of the difference series 10 and the number of the difference series

Three numbers are equal differences, then.

a1+a3=2a2

Therefore the sum of a1, a3 must be even.

Then a1 and a3 are even numbers, and according to the knowledge of permutation, there are a total of a(10,2)=90.

or a1, a3 is an odd number. According to the permutation knowledge, there is a total of a(10,2)=90, so there are a total of 180 equal difference series.

There are 18 tolerances of 1

There are 16 tolerances of 2

There are 14 tolerances of 3

And so on: 12 with a tolerance of 4, 10 with a tolerance of 5, 8 with a tolerance of 6, 6 with a tolerance of 7, 4 with a tolerance of 8, and 2 with a tolerance of 9.

Therefore, the number of such a series of equal differences is 2+4+......18 = (2 + 18) * 9 2 = 90 pcs.

The same negative number is also 90. i.e. 1,2,3 and 3,2,1 are different sequences.

A total of 180 results were found.

The difference is an integer There are 9 groups).

9 groups) 8 groups.

It was 8 groups, 7 groups, 7 groups, 6 groups, 6 groups, 5 groups, 5 groups, 4 groups, 4 groups, 3 groups, 3 groups, 2 groups, 2 groups, 1 group, and 1 group.

The last 1 group is:

So there are a total of 2 (1+2+....)9) 90 groups.

180 pcs., 3rd floor correct. 4, 6th floor is 90, it should be doubled, your tolerance is positive, the same is negative is 90. i.e. 1,2,3 and 3,2,1 are different sequences.

If 123 and 321 are different sequences, it is doubled, 180

Categories: Education Academic Examination >> Learning Help Elder Shouting Problem Description:

Knowing that a4=10, a7=19 of {an}, find the values of a1, d, and s12?

Analysis: Solution: by a7=19;a4=10 Dekai type:

From the property of the difference series an=a1+(n-1)d, we get:

a7=a4+3d=19

10+3d=19

d=3a4=a1+3d

10=a1+9

a1=1 is obtained from the first n terms of the equal difference series and the formula sn=na1+n(n-1)*d 2

s12=12*1+12*(12-1)*3/2=12+198=210

Solution: Let the general formula of the difference series be an an ten b, according to the conditions of the problem, list the binary equations about a and b, and solve the values of a and b, so that the problem is solved.

Answer: The latter item is minus the previous item to make the difference

Again, make the difference: get the difference series with a common ratio of 2.

So 16 is followed by -32

So 17 is followed by 17-32=-15

Equations for the difference series.

Equations for the difference series.

The formula for the difference series is an=a1+(n-1)d

The sum of the first n terms is: sn=na1+n(n-1)d 2 if the tolerance d=1: sn=(a1+an)n 2 if m+n=p+q: am+an=ap+aq, if m+n=2p, then: am+an=2ap

The above n are positive integers.

Text translation. The value of the nth term an = first term + (number of terms - 1) tolerance.

The sum of the first n terms sn=first term + last term Number of terms (number of terms-1) tolerance 2 tolerance d=(an-a1) (n-1).

Number of Items = (Last Item - First Term) Tolerance + 1

When the number column is an odd number, the sum of the first n terms = the number of intermediate terms.

The number column is an even number of terms, find the first and last terms, add the first and last terms, divide it by the sum of 2 equal differences, and the formula for the middle term is 2an+1=an+an+2, where is the equal difference series.

Number of Items = (Last Item - First Term) Tolerance + 1The difference series is a common series, if a series from the second term, the difference between each term and its previous term is equal to the same constant, this series is called the difference series, and this constant is called the tolerance of the difference series, and the tolerance is often denoted by the letter d.

Equations for the difference series. The value of the nth term an = first term + (number of terms - 1) tolerance.

The sum of the first n terms, sn=first term n + number of terms (number of terms 1) tolerance 2.

Tolerance d=(an-a1) (n-1) (where n is greater than or equal to 2 and n is a positive integer).

Number of Items = (Last Item - First Term) Tolerance + 1

Last Item First Item (Number of Items 1) Tolerance.

When the number column is an odd number, the number of the first n terms and the middle terms are the number of terms.

The number column is an even number of terms, and the sum of the first n terms (the sum of the first and last terms) 2.

The formula 2an+1=an+an+2 in the equivariance series, where {an} is the equivariance series.

The sum of the difference series (first term, last term) number of terms 2.

Categories: Education Academic Exams >> Study Help.

Problem description: Set it to an equal difference series, and (a1) 2+(a11) 2=<100, denote s=a1+a2+......A11 then the value range of s is

Analysis: This time it should be right.

Write a6=a, the tolerance is d, then s=11a, (a-5d) 2+(a+5d) 2 100

Simplification yields 2a 2 100-50d 2

Thus a 2 50 (take " " if and only if d = 0).

And because d 2 = 2 has a = 0, and the function f(d) = 100-110d 2 is a continuous function, and decreases with the increase of d.

Therefore, there is a -5*2 (1 2) reserve a 5*2 (1 cavity front 2), so -55*2 (1 2) s=11a 55*2 (1 2) Note] * is the number of the multiply.

Number of Items = (Last Item - First Term) Tolerance + 1

Example: 11 12 13 ....＋31＝？

Analysis and Solution: This string adds 11, 12, 13 ,..., 31 is a series of equal differences, the first term is 11, the last term is 31, there are 31-11 1 21 (terms).

Original = (11 + 31) 21 2 = 441.

When using the equation of equal difference series, sometimes the number of terms is not clear at a glance, so you need to find the number of terms first. According to the relationship between the first term, the last term, and the tolerance, it can be obtained.

Number of Items = (Last Term-First Term) Tolerance + 1, Last Item = First Item + Tolerance (Number of Terms-1).

Finding the number of terms in an equal difference series requires knowing the first term, the last term, and the tolerance, and then calculates it according to the general term formula or summation formula of the equal difference series. Here's how:

If the first term $a 1$, the last term $a n$ and the tolerance $d$ of the equal difference series are known, the number of terms $n$$ can be found by the following formula: $n = frac+1$$, where $n$ denotes the number of terms. Another way to find the number of terms in a series of equal differences is to use the formula for summing the number of terms in a series of equal differences, which represents the general formula for the sum of the first $n$ terms in the series of equal differences: $$s n= frac$$, where $s n$ denotes the sum of the first $n$ terms.

Through the formula for summation of the series of equal differences, the formula for the number of terms $n$ can be further derived: $$n= frac$$ where $s n$ is the sum of the first $n$ terms of the known series of equal differences, and $a 1$ and $a n$ are the first and last terms of the series of equal differences, respectively.

To sum up, to find the number of terms in an equal difference series, you need to know the first term, the last term, and the tolerance, and then calculate it according to the general term formula or summation formula of the equal difference series. The formula for calculating the number of terms can be solved by the first, last and tolerance terms of a known series of equal differences, or the sum of the first $n$ terms of a known series of equal differences.

For example, if we know that the first term of a difference series is $a 1$ and the tolerance is $d$, and all the terms in the difference series are positive integers, then we can quickly estimate the number of terms in the difference series by:

1.Calculate the value of the smallest positive integer term in the sequence, i.e., $a m=a 1+(m-1)d$, where $m$ is a positive integer.

2.Calculate the value of the largest positive integer term in the sequence, i.e., $a n=a 1+(n-1)d$, where $n$ is a positive integer.

3.For a series of equal differences with a starting term of $a 1$ and a tolerance of $d$, if $a m leqslant1 <>

For example, for the first term of the basis hole of an equal difference series is $2$ and the tolerance is $3$, and all the terms in the difference series are positive integers, we can calculate the number of positive integer terms in the difference series as $n=33$ according to the above method.

In addition to the above methods, there are some other methods to solve the number of terms of an equal difference series, such as dichotomy, recursion, etc. However, these methods require a high level of mathematical foundation and computational ability, and are generally only applied in advanced mathematics or related fields.

If you know any two terms, you can find the tolerance and the first term, then an=a1+(n-1)d

Any one can be found according to the general formula.