Three consecutive terms of an arithmetic sequence have a sum of 12 and a product of -80. find the terms. hint: let the terms be x-d, x and x+d.

(x - d) + x + (x + d) = 12 --> Create an equation using the first piece of information - "Three consecutive terms... have a sum of 12"

x - d + x + x + d = 12 --> Simplify the left side of this equation (d cancels out)

3x = 12 --> Divide both sides by 3

x = 4


Use the value of x (x = 4) to find the value of d. To do this, set up another equation using the second piece of information.

(x - d) * (x + d) * x = - 80 --> "Three consecutive terms... have... a product of -80". Then, substitute the value of x (4) into this equation.

(4 - d) * (4 + d) * 4 = - 80 --> Multiply out the sets of brackets, the * 4 is dealt with afterwards

4(16 - 4d + 4d - d²) = - 80 --> Simplify the expression inside the brackets

4(16 - d²) = - 80 --> Multiply out these brackets by the 4

64 - 4d² = - 80 --> Subtract 64 from both sides

- 4d² = - 144 --> Divide both sides by - 4

d² = 36 --> Square root both sides

d = 6

Now, find the values of the terms of the sequence by using substituting the values of x and d into the expressions given.

1. x - d = 4 - 6 = - 2
2. x = 4
3. x + d = 4 + 6 = 10

The three terms are - 2, 4, 10.

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