Question

Determine K so that (3K-2), (4k-6 )
& (K+2) are there consecutive
terms of AP.​

Answer 1

Answer:

(3k-2)=

(k+2)=0

k=-2 put the value

(3.-2-2)=(3×-4)= -12

(4k-6) = (4.-2-6)= (4×-8)= -32

Answer 2

Answer:

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

= 4k - 6 - 3k + 2

= k - 4

also

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

= k+2 - 4k + 6

-3k +8

therefore

k - 4 = -3k +8

4k = 12

k = 3

hope it helps

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