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Q.9. Determine K so that are three consecutive terms of an A.P.
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Dear Student,
Please find below the solution to the asked query:
Given : k2 + 4 k + 8 , 2 k2 + 3 k + 6 and 3 k2 + 4 k + 4 are in AP.
We know In A.P. common difference is equal , So
( 2 k2 + 3 k + 6 ) - ( k2 + 4 k + 8 ) = ( 3 k2 + 4 k + 4 ) - ( 2 k2 + 3 k + 6 )
2 k2 + 3 k + 6 - k2 - 4 k - 8 = 3 k2 + 4 k + 4 - 2 k2 - 3 k - 6
k2 - k - 2 = k2 + k - 2
2 k = 0
So,
k = 0 ( Ans )
Hope this information will clear your doubts about Arithmetic Progressions .
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
Given : k2 + 4 k + 8 , 2 k2 + 3 k + 6 and 3 k2 + 4 k + 4 are in AP.
We know In A.P. common difference is equal , So
( 2 k2 + 3 k + 6 ) - ( k2 + 4 k + 8 ) = ( 3 k2 + 4 k + 4 ) - ( 2 k2 + 3 k + 6 )
2 k2 + 3 k + 6 - k2 - 4 k - 8 = 3 k2 + 4 k + 4 - 2 k2 - 3 k - 6
k2 - k - 2 = k2 + k - 2
2 k = 0
So,
k = 0 ( Ans )
Hope this information will clear your doubts about Arithmetic Progressions .
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards