Question

find 8 consecutive ingeters whose sum is 8028

Answer 1

Answer:

The 8 consecutive numbers are 1000, 1001, 1002, 1003, 1004, 1005, 1006, 1007

Step-by-step explanation:

$x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 + x + 7 = 8028$

$8x + 28 = 8028$

$8x = 8028 - 28$

$\frac{8x}{8} = \frac{8000}{8}$

$x = 1000 \\ x + 1 = 1001 \\ x + 2 = 1002 \\ x + 3 = 1003 \\ x + 4 = 1004 \\ x + 5 = 1005 \\ x + 6 = 1006 \\ x + 7 = 1007$