Consecutive integer help please!?

Question

Find 4 consecutive integers whose sum is twice the cube of 5.

please show the set up and steps to reach the answer

Answer 1

The cube of 5 (5x5x5) is 125. Twice that is 250. Therefore, the problem now is to find four consecutive integers whose sum is 250. That is fairly simple.

A consecutive integer is one greater than the one before it, or x+1. So four consecutive integers could be expressed algebraically as x+x+1+x+2+x+3, or 4x+6.

Then we have:
4x+6=250.
4x=250-6=244
4x=244
x=244/4
x=61.

So the first integer is 61.
Then the others must be 62, 63, and 64. To check, you can add them all together. 61+62+63+64=250.

Problem solved.

Answer 2

Well, I would define what the cube of 5 is. It's 125, so double that would be 250.

Then, we have an equation of (x) + (x + 1) + (x + 2) + (x + 3) = 250. X is the first consecutive digit. Simplify it to 4x+6=250. Subtract 6 from both sides to get 4x=244. X is 61, your first number. So, the four consecutive integers are 61, 62, 63, and 64.

Answer 3

First, find the sum of 5 cubed *2:

5 * 5 * 5 = 125; 125*2 = 250

Next set it up so that x, x+1, x+2, and x+3 add up to 250, then solve for x:

x + x+1 + x+2 + x + 3 =250
4x + 6 = 250
4x = 244
x = 61
x+1 = 62
x+2 = 63
x + 3 = 64

Answer 4

5 cubed is 125

125 times two is 250

250 divided by 4 is 62.5

62.5 is the center of your four integers

therefore, choose 61, 62, 63, 64

61+62+63+64=250

shazaam

Answer 5

(n) + (n+1) + (n+2) + (n+3) = 2(5)^3
4n + 6 = 250
4n = 244
n = 61

So the numbers are 61,62,63 and 64.

Answer 6

x+x+1+X+2+X+3=(5*5*5)*2
4x+6=250
4x=244
x=60
x+1=61
x+2=62
x+3=63