Question 922706: Find the value of k so that 2k + 2, 5k - 11, and 7k - 13 will form an arithmetic progression. Found 2 solutions by srinivas.g, MathTherapy:Answer by srinivas.g(540) (Show Source):
You can put this solution on YOUR website! In arithmetic progression, rule is as follows
multiply with 2 on both sides
5k*2-11*2 = 9k-11
10k-22=9k-11
move 9k to the right
10k-22-9k= -11
k-22 =-11
move -22 to the left
k =-11+22
k =11
result k =11
You can put this solution on YOUR website!
Find the value of k so that 2k + 2, 5k - 11, and 7k - 13 will form an arithmetic progression.
In an arithmetic progression, the common difference, or d is obtained by SUBTRACTING the
1st term from the 2nd term, or by SUBTRACTING the 2nd term from the 3rd term. Thus, we get:
5k – 11 – (2k + 2) = 7k – 13 – (5k – 11)
5k – 11 – 2k – 2 = 7k – 13 – 5k + 11
3k – 13 = 2k – 2
3k – 2k = - 2 + 13
