If (a_n=3n^2+2n-5), what is (a_{n+2}-a_n)?
Answer and explanation
Correct answer: \(12n+16\)
Given \(a_n=3n^2+2n-5\), we get \(a_{n+2}=3(n+2)^2+2(n+2)-5=3n^2+14n+11\). Hence, \(a_{n+2}-a_n=(3n^2+14n+11)-(3n^2+2n-5)=12n+16\). The option \(12n+20\) results from an incorrect subtraction of the constant terms. Exam tip: substitute \(n+2\) into every term first, then combine like terms carefully.
Frequently asked questions
What is the correct answer to this question?
\(12n+16\)
Why is this the correct answer?
Given \(a_n=3n^2+2n-5\), we get \(a_{n+2}=3(n+2)^2+2(n+2)-5=3n^2+14n+11\). Hence, \(a_{n+2}-a_n=(3n^2+14n+11)-(3n^2+2n-5)=12n+16\). The option \(12n+20\) results from an incorrect subtraction of the constant terms. Exam tip: substitute \(n+2\) into every term first, then combine like terms carefully.
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.
