Question

Determine k so that k+2,4k-6 and 3k-2 are three consecutive terms of an ap

Answer 1

Answer:

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

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

3k-8=-k+4

4k=12

k=3

Answer 2

Answer:

Step-by-step explanation:

sorry about the linguistic barrier (I'm Vietnamese so my knowledge about English is limited)

I think that those are three consecutive term of an integer

because k+2 ,4k+6, and 3k-2 are three consecutive term of an integer  

hence, there are two odd number ( because 4k+6 is even number )

We have k+2 <4k-6<3k-2

=> k>8/3 and k<4

=> k =3