If (a_n=2n^2+3n) and (a_n=65), what is the value of (n)?
Answer and explanation
Correct answer: \(5\)
Given \(2n^2+3n=65\), we get \(2n^2+3n-65=0\). Factoring gives \((n-5)(2n+13)=0\), so \(n=5\) or \(n=-\frac{13}{2}\). Since a term number in a sequence must be a positive integer, \(n=5\) is correct. For example, \(n=6\) gives 90, not 65. Exam tip: after solving, always check whether the value is a valid positive integer term number.
Frequently asked questions
What is the correct answer to this question?
\(5\)
Why is this the correct answer?
Given \(2n^2+3n=65\), we get \(2n^2+3n-65=0\). Factoring gives \((n-5)(2n+13)=0\), so \(n=5\) or \(n=-\frac{13}{2}\). Since a term number in a sequence must be a positive integer, \(n=5\) is correct. For example, \(n=6\) gives 90, not 65. Exam tip: after solving, always check whether the value is a valid positive integer term number.
Which subject and chapter does this question cover?
This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
