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If (a_n=15n-8), what is (k) for (a_{6k}-a_{2k}=300)?

Answer and explanation

Correct answer: 5

Given \(a_n=15n-8\), we get \(a_{6k}=15(6k)-8=90k-8\) and \(a_{2k}=15(2k)-8=30k-8\). Therefore, \(a_{6k}-a_{2k}=(90k-8)-(30k-8)=60k\). Using \(60k=300\), we obtain \(k=5\). If \(k=4\), the difference would be only \(240\), so it is not correct. Exam tip: when subtracting two terms of this form, the identical constant terms cancel out.

Related tags

Arithmetic ProgressionNth TermAlgebraic SubstitutionSequence IndicesClass 10 MathematicsFinding The $N$Th Term Of An ApFinding The N Th Term Of An ApArithmetic Progressions (Ap)

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