If (a_n=an+b), (a_2+a_5=29), and (a_3+a_6=39), what will be (a_{10})?
Answer and explanation
Correct answer: 47
Given \(a_n=an+b\), we have \(a_2+a_5=(2a+b)+(5a+b)=7a+2b=29\) and \(a_3+a_6=9a+2b=39\). Subtracting the first equation from the second gives \(2a=10\), so \(a=5\). Substituting in \(7a+2b=29\) gives \(b=-3\). Hence, \(a_{10}=10(5)-3=47\). The nearby option 45 is incorrect because it would require \(b=-5\). Exam tip: subtract such paired-sum equations first, as the constant term often cancels out.
Frequently asked questions
What is the correct answer to this question?
47
Why is this the correct answer?
Given \(a_n=an+b\), we have \(a_2+a_5=(2a+b)+(5a+b)=7a+2b=29\) and \(a_3+a_6=9a+2b=39\). Subtracting the first equation from the second gives \(2a=10\), so \(a=5\). Substituting in \(7a+2b=29\) gives \(b=-3\). Hence, \(a_{10}=10(5)-3=47\). The nearby option 45 is incorrect because it would require \(b=-5\). Exam tip: subtract such paired-sum equations first, as the constant term often cancels out.
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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