Question

Determine k if k+2,4k-6 and 3k-2 are three consecutive term of an A.P.

Answer 1

as k+2, 4k-6, 3k-2 are in AP
∴ the common difference between them is same
⇒ 4k-6-(k+2) = 4k -6-k-2 = 3k-8 ------------(1)
⇒3k -2-(4k-6) = 3k-2-4k+6 = k+4-----------(2)
from 1 and 2
⇒3k-8 = k+4 
⇒3k-k=4+8
⇒2k=12
∴ k= 6


Thank you....!!!
Mark as brainliest if helpfull
Yours, Jahnavi

Answer 2

Answer:

(k+2)+(3k−2)=2(4k−6)

⇒4k=8k−12

⇒4k−8k=−12

⇒−4k=−12

⇒4k=12

⇒k=

4

12

=3

Hence k=3.