If (a_n=7n+c) and (a_6=61), what is (r) when (a_{4r}=299)?
Answer and explanation
Correct answer: 10
Given \(a_n=7n+c\). Substituting \(n=6\), \(a_6=42+c=61\), so \(c=19\). Now \(a_{4r}=7(4r)+19=28r+19\). From \(28r+19=299\), we get \(28r=280\), hence \(r=10\). If 9 were used, the term would be 271, not 299. Exam tip: first find the constant \(c\) from the given term, then substitute the required index in the formula.
Frequently asked questions
What is the correct answer to this question?
10
Why is this the correct answer?
Given \(a_n=7n+c\). Substituting \(n=6\), \(a_6=42+c=61\), so \(c=19\). Now \(a_{4r}=7(4r)+19=28r+19\). From \(28r+19=299\), we get \(28r=280\), hence \(r=10\). If 9 were used, the term would be 271, not 299. Exam tip: first find the constant \(c\) from the given term, then substitute the required index in the formula.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the $n$th term of an AP.
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