Question

Determine k so that (3k-2), (4 k-6) and (k+2) are three consecutive terms of an AP.​

Answer 1

Step-by-step explanation:

(3k -2), (4k-6) and (k+2) are in AP

(4k-6)-(3k-2) = (k+2) - (4k-6)

4k-6-3k+2 =k+2-4k+6

k-4=8-3k

4k = 12

k= 3

Answer 2

SOLUTION:-

Given,

3k-2, 4k-6 & k+2 are three consecutive terms of an A.P.

Therefore,

a,b, c

2b = a+c

b-a = c -b

=) (4k-6) -(3k-2) = (k+2)-(4k-6)

=) 4k-6 - 3k +2 = k+2- 4k +6

=) k-4 = -3k +8

=) k +3k = 8 +4

=) 4k = 12

=) k = 12/4

=) k = 3

Hope it helps ☺️