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.
Frequently asked questions
What is the correct answer to this question?
5
Why is this the correct answer?
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.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
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