If there consecutive terms of an AP are k+2 ,4k -6 and 3k-2 . What is the value of k​

Explanation:

To go from the first term to the second term, we add on some common difference d.

So,

B = A+d

B = (k+2)+d

4k-6 = k+2+d

4k-6-k-2 = d

d = 3k-8

Similarly, to go from the second term to the third term, we also add on d

C = B+d

C = (4k-6)+d

C = (4k-6)+(3k-8)

C = 7k-14

3k-2 = 7k-14

Let's solve for k

3k-2 = 7k-14

-2+14 = 7k-3k

12 = 4k

4k = 12

k = 12/4

k = 3  is the final answer

If k = 3, then we have these three terms:

The arithmetic progression (AP) is 5, 6, 7. The common difference is d = 1.

Note how d = 3k-8 = 3(3)-8 = 1

were is the equation

the answer will be the last one