Question 922706

Find the value of k so that 2k + 2, 5k - 11, and 7k - 13 will form an arithmetic progression.
<pre>
In an arithmetic progression, the common difference, or d is obtained by SUBTRACTING the
1<sup>st</sup> term from the 2<sup>nd</sup> term, or by SUBTRACTING the 2<sup>nd</sup> term from the 3<sup>rd</sup> 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 
{{{highlight_green(highlight_green(k = 11))}}}