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  • Determine k so that k^2+ 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4 are three consecutive terms of an AP.

Determine k so that  k^2+ 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4 are three consecutive terms of an AP.

Answers (1)

Here the given AP is 

 k\textsuperscript{2} + 4k + 8, 2k\textsuperscript{2} + 3k + 6, 3k\textsuperscript{2} + 4k + 4\\

a\textsubscript{1}= k\textsuperscript{2} + 4k + 8, a\textsubscript{2}=2k\textsuperscript{2}+3k+6, a\textsubscript{3}=3k\textsuperscript{2} + 4k + 4\\

\\a\textsubscript{2}- a\textsubscript{1} = 2k\textsuperscript{2} + 3k + 6 - (k\textsuperscript{2} + 4k+ 8)\\ = 2k\textsuperscript{2} + 3k + 6 - k\textsuperscript{2} - 4k - 8\\ = k\textsuperscript{2} - k - 2\\ a\textsubscript{3} - a\textsubscript{2} = 3k\textsuperscript{2} + 4k + 4 - (2k\textsuperscript{2}+3k + 6) \\ = 3k\textsuperscript{2} + 4k + 4 - 2k\textsuperscript{2} - 3k - 6\\ = k\textsuperscript{2} + k - 2\\

 Hence, it is given that these terms are an AP 

 So the common difference of the A.P  

\\a\textsubscript{2} - a\textsubscript{1} = a\textsubscript{3} - a\textsubscript{2}\\ k\textsuperscript{2} - k - 2 = k\textsuperscript{2} + k - 2\\ k\textsuperscript{2} - k - 2 - k\textsuperscript{2} - k + 2 = 0\\ -2k = 0\\ k = 0\\

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infoexpert21

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