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

Answers

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 =  - 2

The 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

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

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

-4k+5=2k-13

13+5=2k+4k

18=6k

18÷6=6k÷6

3=k

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
The value of k is 3 po.


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