The sum of three consecutive odd numbers is 57. Can you use the equation to find these three numbers

The sum of three consecutive odd numbers is 57. Can you use the equation to find these three numbers

Let the number in the middle be x, and the three numbers are X-2, x, x + 2
x-2+x+x+2=57
3x=57
x=19
The three numbers are 17, 19 and 21 respectively

There is an equation!; the sum of three consecutive odd numbers is 57, take these three numbers

Let the middle odd number be x, then the three odd numbers are: X-2, x, x + 2
There is an equation
x-2+x+x+2=57
The solution is x = 19
So these three numbers are 17, 19, 21

The sum of three consecutive odd numbers is 75______ ．

Let the smallest odd number be x, then the others are x + 2, x + 4  x + X + 2 + X + 4 = 75. The solution is: x = 23, the three numbers are 23, 25 and 27 respectively

Given that the sum of 3 consecutive odd integers in 5 consecutive integers is 15 more than the sum of 2 even integers, find these 5 consecutive integers
This is a problem of making a linear equation of one variable

Let the first number be x, then all the other numbers will be known. They are x plus 1,2,3,4. Just list the equations according to the meaning of the question. It's easy to solve X. note that x is an odd number. Pay attention to the choice. The answer is very simple

If the sum of three consecutive odd integers in five consecutive integers is 15 more than the sum of two even integers, find the five consecutive integers

The sum of three consecutive odd integers in five consecutive integers is 15 times more than the sum of two even integers
These five consecutive integers
2n-3,2n-2,2n-1,2n,2n+1
(2n-3)+(2n-1)+(2n+1)-(2n-2)-2n=15
6n-3-4n+2=15
2n-1=15
2n=16
n=8
These five consecutive integers are 13, 14, 15, 16 and 17

It is known that there are 10 items in the arithmetic sequence, in which the sum of odd items is 15 and the sum of even items is 30, then the tolerance is?

Tolerance that the sum of even items minus the sum of odd items is 5 times
(a2-a1)+(a4-a3)+(a6-a5)+（a8-a7）+（a10-a9）
=(a2+a4+a6+a8+a10)-(a1+a3+a5+a7+a9)
=5d
That is 30-15 = 15 = 5D
d=3

The sum of even numbers of five continuations is 180, the number in the middle is y, what is y

The middle number is y, and the five numbers are continuous even numbers, so the first two digits are y-4, Y-2
Similarly, the following two numbers are y + 2, y + 4
The sum of 5 numbers is 180
therefore
Y-4+Y-2+Y+Y+2+Y+4=180
Y=36
The answer is y

The sum of three consecutive even numbers is 60. What's the number in the middle

Let the middle number be x, then the minimum number is (x - 2) and the maximum number is (x + 2)
Formula:
x+（x+2）+（x-2）=60
3x=60
x=20

It is known that the sequence {an} satisfies A1 = 1, an + 1 = {1 / 2An + N, n is odd, an-2n, n is even}
(1) Find A1, A2;
(2) When n > 2, find the relationship between a2n-2 and A2N, and find the general formula of even number terms in the sequence {an};
(3) Finding the sum of all odd numbers in the first 100 items of the sequence {an}

1a1=1,a2=1/2a1+1=3/2
2. When n is an even number ＞ 2, an = 1 / 2A (n-1) + (n-1) = 1 / 2 [a (n-2) - 2 (n-2)] + n-1 = 1 / 2A (n-2) + 1
So let n = 2K (K ∈ z), a2k = 1 / 2A2 (k-1) + 1
So a2k-2 = 1 / 2A2 (k-1) - 1 = 1 / 2 [A2 (k-1) - 2]
The {a2k-2} is an equal ratio sequence with the first term of - 1 / 2 and the common ratio of 1 / 2
a2k-2=-1/2*(1/2)^(k-1)=-(1/2)^k
So the general formula of even term in {an} is an = - (1 / 2) ^ (n / 2) + 2 (n is even)
3. N is odd, an = (a (n + 1) - n) * 2 = {- (1 / 2) ^ [(n + 1) / 2] + 2-N} * 2 = - (1 / 2) ^ [(n-1) / 2] - 2 (n-2)
a1+a3+...+a99=-1-(1/2)^1-(1/2)^2-...-(1/2)^49-2-6-10-...-198+4*50=
-(1-(1/2)^50)/(1-1/2)-(2+198)*50/2+200=-4802+2^(-49)

It is known that the sequence {an} satisfies A1 = 1, an + 1 = 1 / 2An + n (when n is odd), an + 1 = an-2n (when n is even), and bna2n-2, n ∈ n*
Prove that the sequence {BN} is an equal ratio sequence and find its general term formula
It's the front, with brackets. sorry

Please confirm whether it is an + 1 = 1 / 2 (an + n) or (1 / 2An) + n; - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -