The value of k so that 5k-3, k+2, and 3k-11 will form an arithmetic sequence is 3.

Step-by-step explanation:

First, find the difference between the consecutive terms. In subtracting polynomials, change the sign of each term of subtrahend, then proceed to addition rule,

The sequence:

5k-3, k+2, 3k-11

a) (k+2) - (5k-3)

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

= -4k + 5

b) (3k-11) - (k+2)

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

= 2k - 13

Equate the result (difference between the terms):

-4k + 5 = 2k - 13

-4k - 2k = -13 -5

-6k = -18

-6k/-6 = -18/-6

k = 3

ANSWER: The value of k is 3.

Check:

5k - 3 = 5(3) - 3 = 15 - 3 = 12k + 2 = (3) + 2 = 5 3k - 11 = 3(3) - 11 = 9 - 11 = - 2The arithmetic sequence:

12, 5, -2

The difference between consecutive terms should be common/equal:

-2 - 5 = -7

5 - 12 = -7

The sequence is decreasing, and the difference between terms is -7.

First CD = ( k+2) - (5k-3) = k + 2 - 5k + 3 = -4k + 5

second CD = (3k-11) - (k+2) = 3k-k -11 -2 = 2k-13

thus

-4k + 5 = 2k-13

5 + 13 = 6k

18 = 6k

k = 3

The 3 expressions form an arithmetic sequence with - 7 as the CD

D1=k+2-(5k-3)

D1=k+2-5k+3

D1=-4k+5

D2=3k-11-(k+2)

D2=3k-11-k-2

D2=2k-13

D1=D2

-4k+5=2k-13

5+13=2k+4k

18=6k

Divide both side by 6

K=3

5k-3, k+2, 3k-11

a1 a2 a3

a3-a2=a2-a1

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

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

2k-13=-4k+5

2k+4k=5+13

(6k=18)/6

k=3

5(3)-3=12

3+2=5

3 (3)-11=-2

5k-3, k+2, 3k-11

12 , 5 , -2

1. Determine first the common difference.

A5 = A2 + (4- 1) d

3 = 24 + (4 - 1) d

3 = 24 + 3d

3-24 = 3d

-21 = 3d

-21/3 = 3d/3

d = -7

Next, get the first term.

An = A1 + (n-1) d

3 = A1 + (5-1) -7

3 = A1 + (4) -7

3 = A1 + (-28)

3 + 28 = A1

A1 = 31

A5 = A2 + (4- 1) d

3 = 24 + (4 - 1) d

3 = 24 + 3d

3-24 = 3d

-21 = 3d

-21/3 = 3d/3

d = -7

Next, get the first term.

An = A1 + (n-1) d

3 = A1 + (5-1) -7

3 = A1 + (4) -7

3 = A1 + (-28)

3 + 28 = A1

A1 = 31