Question

for what value of k that k - 1, 3 k + 1 and 16 K + 2 will be in GP

Answer 1

k -1 , 3k + 1 and 16k + 2 are in GP
so, it will follow the condition of GP.
e.g , common ratio of terms = constant

so, (3k + 1)/(k -1) = (16k +2)/(3k + 1)
(3k +1)² = (k -1)(16k +2)
9k² + 6k +1 = 16k² + 2k -16k -2
7k² - 20k - 3 = 0
7k² -21k + k -3 = 0
(7k + 1)(k -3) = 0
k = -1/7 and 3

Answer 2

take k-1 to be t1
        3k+1 to be t2
       16k+2 to be t3
 t2-t1=t3-t2
3k+1-k+1=16k+2-3k-1
2k+2=13k+1
2k+1=13k
11k=1
k=1/11