If (a_n=n^2+10), which term is (131)?
Answer and explanation
Correct answer: 11th
To find which term has value 131, set \(a_n=131\): \(n^2+10=131\). Thus, \(n^2=121\), so \(n=11\). Since a term number is a positive integer, 131 is the 11th term of the sequence. The 12th term is \(12^2+10=154\), so it is not correct. Exam tip: To find the position of a given term, equate \(a_n\) to the given value first.
Frequently asked questions
What is the correct answer to this question?
11th
Why is this the correct answer?
To find which term has value 131, set \(a_n=131\): \(n^2+10=131\). Thus, \(n^2=121\), so \(n=11\). Since a term number is a positive integer, 131 is the 11th term of the sequence. The 12th term is \(12^2+10=154\), so it is not correct. Exam tip: To find the position of a given term, equate \(a_n\) to the given value first.
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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