Hello, I have been trying a few ways to do this problem and am not having success.
The sum of four consecutive odd integers is -212. Write an expression for the second, third and fourth odd integers. Use these expressions to write, and solve an equation to find the four integers.
So, I tried first:
n + n+2 + n+4 + n+6 = 212
my answer -56, which is even so no good.
then i tried
2n+1 + 2n+3 + 2n+5 + 2n+7 = -212
my answer came out to -28.5
which is a decimal, so no good.
Can you please help?
The data is flawed.
4 consecutive ODD numbers can not add up to -212.
They must be consecutive EVEN integers.
Your answer of -56 is correct.
notice
-56-54-52-50 = -212
Thanks! That's what I thought
Of course! Solving this problem involves setting up an equation correctly and applying the concept of consecutive odd integers.
Let's start by defining the first odd integer as "n." Since we are asked to write expressions for the second, third, and fourth odd integers, we need to understand the pattern of consecutive odd integers.
An odd integer is represented as 2k + 1, where k is an integer. For consecutive odd integers, we can use the pattern:
1st odd integer: n
2nd odd integer: n + 2
3rd odd integer: n + 4
4th odd integer: n + 6
Now, we can set up an equation using the given information: the sum of these consecutive odd integers is -212:
n + (n + 2) + (n + 4) + (n + 6) = -212
To solve this equation, we can combine like terms and simplify:
4n + 12 = -212
4n = -224
n = -56
So, the first odd integer, n, is -56.
Using our expressions for the second, third, and fourth odd integers, we can substitute n = -56:
2nd odd integer: -56 + 2 = -54
3rd odd integer: -56 + 4 = -52
4th odd integer: -56 + 6 = -50
Therefore, the four consecutive odd integers that sum up to -212 are -56, -54, -52, and -50.