MathematicsIf 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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