How do you solve 4 consecutive odd integers with the sum of 184?

If the first odd number is a then the next consecutive odd number is a + 2, the one after that is a + 4 and the one after that is a + 6
(Note a + 1, a + 3 and a + 5 are even numbers)

a + (a + 2) + (a + 4) + (a + 6) = 184
4a + 12 = 184
4a = 172
a = 172/4
a = 43
numbers are 43, 45, 47 and 49

Solving Consecutive Integers

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1. 15=4.2x x=15/4.2 (its an irrational decimal) 2. x=first integer x+2=second integer x+4=third integer x+6=fourth integer x+x+2+x+4+x+6=184 4x+12=184 4x=172 x=43 so the integers are 43, 45, 47, and 49. 3. im not sure if there ISN'T supposed to parentheses around the "2w+2" but i dont think there is.... 2w+2(3w+6)=24 2w+6w+12=24 8w+12=24 8w=12 w=12/8=3/2 4. that isnt an equation so i cant solve it, fix it and i can. sorry

(2k + 1) + (2k + 3) + (2k + 5) + (2k + 7) = 184
8k + 16 = 184
8k = 168
k = 21

Numbers are 43 , 45 , 47 , 49

Let:
A = the first number
k =interval = 2 for consecutive odd numbers( also for consecutive even numbers)
n = number of terms = 4
An = the nth term
Sn = sum of terms

An = A + k(n - 1) = A + 2(4 -1) = A + 6

Sn = (n/2)( A + An) = (4/2)(A + A +6) = 184

2A + 6 = 92

A = (92 - 6)/2 = 86/2 = 43

second term is;

A2 = 43 + 2(2-1) = 45

Third term is;

A3 = 43 + 2(3-1) = 47

Fourth term is;

A4 = 43 + 2(4-1) = 49

w - 2 = x (solve by using substitution)
x - 2 = y
y - 2 = z
w + x + y + z = 184

w - 2 = x
w = x + 2

w + x + y + z = 184
(x + 2) + x + y + (y - 2) = 184
x + 2 + x + y + y - 2 = 184
x + x + 2y = 184 - 2 + 2
2x + 2y = 184
2x + 2(x - 2) = 184
2x + 2*x - 2*2 = 184
2x + 2x - 4 = 184
4x = 184 + 4
4x = 188
x = 188/4
x = 47

w - 2 = x
w - 2 = 47
w = 47 + 2
w = 49

x - 2 = y
47 - 2 = y
y = 45

y - 2 = z
45 - 2 = z
z = 43

∴ w = 49 , x = 47 , y = 45 , z = 43

You make an equation

184 = x + (x+2) + (x+4) + (x+6)
184 = 4x + 12
172 = 4x
x = 43

so 43, 45, 47, 49

Because you need four numbers, divide 184 by 4.
184/4=46
One number smaller and larger than 46 is 45 and 47. This gives you the first 2 odd integers.
For the second two, repeat on 45 and 47 but make sure the number is 2 smaller and 2 larger so the number remains odd.
Answer= 43, 45, 47, 49

Let the integers be x, (x + 2) + (x + 4), (x + 6)
Sum = 4x + 12 = 184
4x = 172
x = 43
integers are 43, 45, 47, 49
-----------

first number: x
second: x+2
third: x+4
Last: x+6

The sum of all is equal to 184:

x+x+2+x+4+x+6+ = 184

So, 4x = 184-12

x=172/4 = 43

the numbers are: 43, 45, 47 and 49 :)