Question

find the value of k so that k+2,4k-6and3k-2 are the three consecutive terms of an A.P​

Answer 1

Step-by-step explanation:

a(1) = k+2 , a(2) = 4k-6

d = a(2) - a(1)

d = 4k-6-k-2

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

a(3) = a+ 2d

3k-2 = (k+2) + 2d

2d = 2k - 4

d = k-2 ----(2)

from equation (1) & (2)

3k - 8 = k - 2

2k = 6

k = 3