What must be the value of k so that 5k-3,k+2, and 3k-11 will form an arithmetic sequence?

Question

arithmetic

Answer 1

In an arithmetic sequence, the difference between any two successive terms is  the same and equal to the common difference d:

 

a_n = a_n-1 + d

a_n- a_n-1 = d

 

Hence:

 

a₃ - a₂ - d

a₂ - a₁ = d

 

Since both equations equal d, set them equal:

 

a₃ - a₂ = a₂ - a₁

(3k-11) - (k+2) = (k+2) - (5k-3)

2k - 13 = -4k + 5

 

Solve for k.

Answer 2

Let d = common difference and a₁ = first term

 

Then, k+2 = a₁ + d and 3k-11 = a₁ + 2d

 

So, k+2 = 5k-3+d and 3k-11 = 5k-3+2d

 

Simplify to get the following system of equations:

 

    4k+d = 5

    2k+2d = -8

 

Divide the second equation by -2:   4k+d = 5

                                                    -k-d = 4

 

Add the equations to obtain 3k = 9

 

                                          k = 3