Find four consecutive odd integers whose sum is -112.

Question

 Can you please help me?

 

Answer 1

2x+3, 2x+1, 2x-1, 2x-3 can be consecutive odd integers if for all x ∈I

which yields a sum of -112, i.e.,  2x+3+2x+1+2x-1+2x-3=-112

=>8x=-112

dividing both sides by 8 we get, x=-14, hence the consecutive numbers are 

2x+3=2*(-14)+3=-28+3=-25

2x+1=2*(-14)+1=-28+1=-27

2x-1=2*(-14)-1=-28-1=-29

2x-3=2*(-14)-3=-28-3=-31

Answer 2

n represents the first odd integer

n + 2 represents the second odd integer

n + 2 + 2 represents the third odd integer

n + 2 + 2 + 2 represents the fourth odd integer

 

n + n + 2 + n + 2 + 2 + n + 2 + 2 + 2 = -112   combine

4n +12 = -112

4n = -124

n = -31

 

What are the other three odd integers?

 

 

 

Answer 3

The integers are consecutive and odd so:

x+(x+2)+(x+4)+(x+6)=-112

4x+12=-112

4x=-124

x=-31

 

So the integers are -31,-29,-27, and -25

 

Proof: 

-31-29-27-25=-112

-112=-112