Question 830939
1.  Based on the statement of the problem, we must have:  b = a + 2; c = a + 4; d = a + 6

2.  Based on the statement of the problem, a + a + 4 = (a + 2)(a + 6)/5

3.  That is, 2a + 4 = (a^2 + 2a + 6a + 12)/5

4.  Multiply the equation above by 5:  10a + 20 = a^2 + 8a + 12

5.  Rearranging, we get a^2 - 2a - 8 = 0

6.  Factorizing, (a - 4)(a + 2) = 0

7.  The solutions to the above are:  a = 4; a = -2

8.  We can ignore the second solution, since a must be a natural number (that is, an integer and greater than 0)

9.  So a = 4; b = 6; c = 8; d = 10

10.  Let us check:  a + c = 4 + 8 = 12;  b times d = 6 times 10 = 60, which is 5 times (a + c)