The sum of five consecutive even integers is 1310. Find the integers

Step-by-step explanation:

Let the first of these even numbers = n

The second one is n + 2

The third one is n + 4

The fourth one is n + 6

The fifth one is n + 8

Their sum is 1310

Equation

n + n + 2 + n + 4 + n + 6 + n + 8 = 1310

Solution

There are 5 ns.

The total of the numbers is 2 + 4 + 6 + 8 = 20

Simplify the given equation

5n + 20 = 1310 Subtract 20 from both sides.

5n + 20 - 20 = 1310 - 20 Combine

5n = 1290 Divide by 5

5n/5 = 1290/5 Divide

Answers

n1 = 258

n2 = 260

n3 = 262

n4 = 264

n5 = 266

Check

The total is 258 + 260 + 262 + 264 + 266 = 1310

It checks.

5 consecutive even integers: n, n+2, n+4, n+6, n+8

n + (n+2) + (n+4) + (n+6) + (n+8) = 1310

combine like terms

5n+20 = 1310

subtract 20 from each side

5n = 1290

divide by 5

n=258

n+2 = 260

n+4=262

n+6=264

n+8=266

Answer: 258,260,262,264,266