If (a_n=2n^2+n-1), what is the value of (a_6-a_2)?
Answer and explanation
Correct answer: 68
Given \(a_n=2n^2+n-1\), \(a_6=2(6)^2+6-1=72+6-1=77\) and \(a_2=2(2)^2+2-1=8+2-1=9\). Therefore, \(a_6-a_2=77-9=68\). The value 72 is only \(2(6)^2\), so it ignores the \(+6-1\) part of the rule. Exam tip: calculate each required term separately before finding their difference.
Frequently asked questions
What is the correct answer to this question?
68
Why is this the correct answer?
Given \(a_n=2n^2+n-1\), \(a_6=2(6)^2+6-1=72+6-1=77\) and \(a_2=2(2)^2+2-1=8+2-1=9\). Therefore, \(a_6-a_2=77-9=68\). The value 72 is only \(2(6)^2\), so it ignores the \(+6-1\) part of the rule. Exam tip: calculate each required term separately before finding their difference.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: Explicit or general rule.
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