Determine k so that k+2, 4k-6 and 3k-2 are three consecutive terms of A.P.
Answers
Answer:
If k+2, 4k-6 and 3k-2 are in A.P , then they would have the same common difference,
4k - 6 - [ k + 2 ] = 3k - 2 - [ 4k - 6]
4k - 6 - k - 2 = 3k - 2 - 4k + 6
3k - 8 = -k + 4
3k + k = 4 + 8
4k = 12
k = 12/4
k = 3
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You can verify if the no:s are in A.P , if k = 3
k + 2 = 5
4k - 6 = 6
3k - 2 = 7
5,6,7 .... are in A.P
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Therefore ,
k = 3
Answer:
If k+2, 4k-6 and 3k-2 are in A.P , then they would have the same common difference,
4k - 6 - [ k + 2 ] = 3k - 2 - [ 4k - 6]
4k - 6 - k - 2 = 3k - 2 - 4k + 6
3k - 8 = -k + 4
3k + k = 4 + 8
4k = 12
k = 12/4
k = 3
===================================
You can verify if the no:s are in A.P , if k = 3
k + 2 = 5
4k - 6 = 6
3k - 2 = 7
5,6,7 .... are in A.P
===========================
Therefore ,
k = 3