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Q.

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


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a

2

b

3

c

4

d

5 

answer is B.

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Detailed 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(4k−6)=k+2+3k−2
⇒8k−12=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
⇒8k−4k=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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