Question

Determine the value of 'k' if k+2,4k-6,3k-2 are three consecutive terms of an AP. ....don't spam. .....or else your answer will be reported

Answer 1

same common difference in A.P.
then,
4k-6–(k+2) = 3k-2–(4k-6)

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

3k-8= –k+4

4k = 12

k =12/4

k=3

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Answer 2

Hii !!☺

The common difference of terms are same if k+ 2 ,4K - 6 and 3 k - 2 are in AP .

4K - 6 - (K + 2)= 3 K - 2 - (4k- 6)
4K - 6 - K - 2 = 3 K - 2 - 4k+ 6
3K - 8 = - K + 4
4K = 12
k = 3

hence the value of k is 3 .


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