In an AP, (a_3+a_8=76) and (a_{13}=92). What is (a_{27})?
Answer and explanation
Correct answer: 192.8
Let the first term be \(a\) and the common difference be \(d\). Then \(a_3+a_8=(a+2d)+(a+7d)=2a+9d=76\), while \(a_{13}=a+12d=92\). Multiplying the second equation by 2 gives \(2a+24d=184\). Subtracting the first equation gives \(15d=108\), so \(d=7.2\). Hence \(a=92-12(7.2)=5.6\), and \(a_{27}=a+26d=5.6+26(7.2)=192.8\). The nearby distractor 196 does not satisfy the given conditions. Exam tip: first convert every stated AP term or sum into equations in \(a\) and \(d\).
Frequently asked questions
What is the correct answer to this question?
192.8
Why is this the correct answer?
Let the first term be \(a\) and the common difference be \(d\). Then \(a_3+a_8=(a+2d)+(a+7d)=2a+9d=76\), while \(a_{13}=a+12d=92\). Multiplying the second equation by 2 gives \(2a+24d=184\). Subtracting the first equation gives \(15d=108\), so \(d=7.2\). Hence \(a=92-12(7.2)=5.6\), and \(a_{27}=a+26d=5.6+26(7.2)=192.8\). The nearby distractor 196 does not satisfy the given conditions. Exam tip: first convert every stated AP term or sum into equations in \(a\) and \(d\).
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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