If k+2,4k−6,3k−2 are the three consecutive terms of an A.P then the value of k is:

# If k+2,4k−6,3k−2 are the three consecutive terms of an A.P then the value of k is:

1. A
2
2. B
3
3. C
4
4. D
5

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### Solution:

We are given that the three terms of an A.P as k+2,4k−6,3k−2
Now, let us use the standard condition of an A.P
We know that the condition of an A.P that is a,b,c the three terms of an A.P then
2b=a+c
By using the above condition to given three terms of the A.P we get
2(4k6)=k+2+3k2
8k12=4k
Now, let us interchange the terms in such a way that the variable terms comes one side and constant terms on the other side then we get
8k4k=12
4k=12
k=3
Therefore we can conclude that the value of k is 3.
So, option (2) is the correct answer.

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