# If (k – 2, (2k +1), (6 k + 3) are in GP, then k = ?

**Answer: (2) 7 **

**Solution:-**

(k – 2), (2k + 1), (6k + 3)

**Formula-
b ^{2} = ac , a = first term, b = second term and c = third term.**

Then,

(2 k + 1)^{2} = (k – 2) (6 k + 3)

4 k^{2} + 1 + 2 x 2 k x 1 = 6k x k + 3 k – 12 k – 6

4 k^{2} + 4 k + 1 = 6 k^{2} – 9 k – 6

2 k^{2} – 13 k – 7 = 0

2 k^{2} – 14 k + k – 7 = 0

2 k ( k – 7) + 1 (k – 7) = 0

(2 k + 1) (k – 7) = 0

k – 7 = 0

k = 7

Then, the value of k is 7.

Hence, the correct answer is option (2) 7.