The sum of four consecutive numbers in an A. P. is 28. The product of the second and third numbers exceeds that of the first and last by 18. Find the
numbers.

Let x be the first number in this sequence of four. Then the next three are x + n, x + 2n, and x + 3n, because consecutive terms in an arithmetic progression differ by a fixed constant.

The sum of these four numbers is 28:

x + (x + n) + (x + 2n) + (x + 3n) = 4x + 6n = 28

or

2x + 3n = 14

The product of the second and third numbers is greater than the product of the first and last numbers by 18:

(x + n)(x + 2n) = x(x + 3n) + 18

Expanding everything on both sides gives

x² + 3nx + 2n² = x² + 3nx + 18

and terms cancel so we end up with

2n² = 18  ==>  n² = 9  ==>  n = 3 or n = -3

If n = 3, then

4x + 6*3 = 28  ==>  4x = 10  ==>  x = 5/2

If n = -3, then

4x + 6*(-3) = 28  ==>  4x = 46  ==>  x = 23/2

So we have two possible solutions, but the only difference between them is the order. If n = 3, then the four numbers are 5/2, 11/2, 17/2, and 23/2. If n = -3, then the four numbers are, again, 23/2, 17/2, 11/2, and 5/2.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116