Find x so that the values given are consecutive terms of an arithmetic sequence. 11+2x, 20-5x, 21+4x

Answer 1

Each consecutive term is the same distance from the previous.

So if we take the second term a₂ and subtract the first term a₁ from it, we should find "d" the common difference. Likewise, a₃ - a₂ should give us the same "common difference" as a₂ - a₁ gave us. Since each of these subractions should give us the same number "d", we can set the results of each subtraction to be equal to each other.

a₁ = 11 + 2x

a₂ = 20 - 5x

a₃ = 21 + 4x

d = a₂ - a₁ = 20 - 5x - (11 + 2x) (distribute the minus sign into the parenthesis)

d = 20 - 5x - 11 - 2x

d = 9 - 7x

also

d = a₃ - a₂ = 21 + 4x - (20 - 5x) (distribute the minus sign into the parenthesis)

d = 21 + 4x - 20 + 5x

d = 1 + 9x

Since both of these subractions = "d" we set them equal to each other:

9 - 7x = 1 + 9x

subtract 1 from both sides

8 - 7x = 9x

add 7x to both sides

8 = 16x

divide both sides by 16

1/2 = x

Let's check.

a₁ = 11 + 2x = 11 + 2(1/2) = 11 + 1 = 12

a₂ = 20 - 5x = 20 - 5(1/2) = 20 - 2.5 = 17.5

a₃ = 21 + 4x = 21 + 4(1/2) = 21 + 2 = 23

Check the common difference: 23 - 17.5 = 5.5, 17.5 - 12 = 5.5

Looking at it differently

a₁ = 12

a₁ + 5.5 = 17.5 = a₂

a₂ + 5.5 = 23 = a₃

_{So x = 1/2}