Question

find k if 4k+8 , 2k^2 + 3k + 6 , 3k^2 + 4k + 4 are three consecutive terms of an AP?

Answer 1

If a,b and c are in AP,

than they both have same common difference,

means ,

b - a = c - b = common difference (d)

As per the question,

$2k^{2} +3k+6 - (4k+ 8) = 3k^{2} +4k+4 - (2k^{2} +3k+6) \\ \\2k^{2} +3k +6-4k-8 = 3k^{2} +4k+4 - 2k^{2} -3k-6 \\\\2k^{2} -k -2 = k^{2} + k -2\\\\2k^{2} - k^{2} = 2k\\\\k^{2} = 2k \\\\k = 2$

Thus the value of K is 2.

Answer 2

Answer:

K = 2

Step-by-step explanation:

hole it's helpful........