Question

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

Answer 1

Step-by-step explanation:

by using the condition to be in AP

i Havel solved it

hope the attachment is useful

Answer 2

Conditions for AP is 2b=a +c

a=k+2 b=4k-6 c=3k-2

2b=a+c

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

8k-12 =4k^2+6k-2k-4

4k^2+6k-2k-8k+12-4=0

4k^2-4k+8=0

(Div the eq. by 4)

k^2-k+2=0

(k-2)(k+1)=0

k=2 or k=-1