If (a_n=2n^2-3n+4), what is (a_{n+1}-a_n)?
Answer and explanation
Correct answer: \(4n-1\)
Given \(a_n=2n^2-3n+4\), replace \(n\) by \(n+1\): \(a_{n+1}=2(n+1)^2-3(n+1)+4=2n^2+n+3\). Therefore, \(a_{n+1}-a_n=(2n^2+n+3)-(2n^2-3n+4)=4n-1\). The option \(4n+1\) can result from an error while subtracting the constant terms. Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).
Frequently asked questions
What is the correct answer to this question?
\(4n-1\)
Why is this the correct answer?
Given \(a_n=2n^2-3n+4\), replace \(n\) by \(n+1\): \(a_{n+1}=2(n+1)^2-3(n+1)+4=2n^2+n+3\). Therefore, \(a_{n+1}-a_n=(2n^2+n+3)-(2n^2-3n+4)=4n-1\). The option \(4n+1\) can result from an error while subtracting the constant terms. Exam tip: write \(a_{n+1}\) separately before subtracting \(a_n\).
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.
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