Write out that four consecutive numbers are their sum is 22, how do you calculate it?

The sum of the four consecutive numbers is 22", that is, the sum of the 4 terms and the difference series with tolerance (d) of 1 is 22

Suppose the first term is A1 and the last term is A4, yes.

sn=(a1+a4)n 2 22=(a1+a4)4 2a4=a1+(4-1)d gives a4=a1+3 and a1=4

by an=a1+(n-1)d

These four numbers are:

Let these 4 numbers be x-1, x, x+1, x+2

x-1+x+x+1+x+2=22

4x+2=22

4x=20x=5

So for 4, 5, 6, 7

Since it is a continuous number, let one of them be an unknown number x, then the remaining three can become an equation with x.

Then add them up to get 22It's easy to calculate.

The next 2 numbers are 6 and 5, and the next number is 7 and 4, so 4 numbers are

Set x. x+(x+1)+(x+2)+(x+3)=22。。。

The solution is x=4, so they are ,

The four consecutive numbers are , and the sum of the consecutive numbers is 22

Let the number in the middle be x, the column equation.

x*(x+1)+x*(x-1)+(x-1)*(x+1)=242 is simplified to 3x 2-1=242

The solution is x=9, so the three numbers are 8, 9, and 10

242 3 is approximately equal to 81

So the three numbers are 8, 9, 10

Let the first integer be x, the second be x+1, and the third x+2 then there is x(x+1)+x(x+2)+(x+1)(x+2)=242, i.e., x 2+x+x 2+2+x+x 2+3x+2=2423x2+6x+2=242

3x^2+6x-240=0

x^2+2x-80=0

x+10)(x-8)=0

x = -10 or 8

So these three numbers are -10 , -9, -8 or for, 8, 9, 10, if you don't understand something, you can ask me, and if you are satisfied, please adopt it. Thanks!

Let these three numbers be x-1 , x , x+1

x-1)x+x(x+1)+(x-1)(x+1)=242x^2-x+x^2+x+x^2-1=2423x^2=243

x^2=81

x=9 or x=-9

So 8, 9, 10 or -10, -9, -8

Let these three numbers be a, b, c, b=a+1, c=a+2, then the old and void ab+ac+bc=242

a(a+1)+a(a+2)+(a+1)(a+2)=242a^2+2a-80=0

a=-10,a=8

Therefore, the three numbers containing the sedan can be -8, -9, and -10

These four numbers are n, n+1, n+2, n+3

n+n+1+n+2+n+3=25

4n+6=25

There is no such thing as a four-consecutive integer addition like 4n=19.

Unlikely, the sum of four numbers in a row must be even.

You have to limit it to an integer.

Consider assumming that the first integer is x, then.

x+(x+1)+(x+2)+(x+3)=18 to get x=34 consecutive integers is: 3, 4, 5, 6

Actually, 18 4=, then we can mentally calculate 3,4,5,6.

Let the first number be x, then the second number is x+1, the third number is x+2, and the fourth number is x+3, then x+(x+1)+(x+2)+(x+3)=18, that is, 4x+6=18, the solution is x=3, so these four consecutive numbers are 3, 4, 5, 6