In an AP (a_6+a_{13}=211) and (a_{22}=256). What is (a_{44})?
Answer and explanation
Correct answer: \(\frac{13022}{25}\)
Let the first term be \(a\) and the common difference be \(d\). Then \(a_6+a_{13}=(a+5d)+(a+12d)=2a+17d=211\), while \(a_{22}=a+21d=256\). Doubling the second equation and subtracting the first gives \(25d=301\), so \(d=\frac{301}{25}\). Hence \(a=256-21d=\frac{79}{25}\). Therefore, \(a_{44}=a+43d=\frac{79}{25}+43\cdot\frac{301}{25}=\frac{13022}{25}\). The value \(539\) does not follow because the given data do not give \(d=14\). Exam tip: write \(a_n=a+(n-1)d\) carefully, especially when converting term indices.
Frequently asked questions
What is the correct answer to this question?
\(\frac{13022}{25}\)
Why is this the correct answer?
Let the first term be \(a\) and the common difference be \(d\). Then \(a_6+a_{13}=(a+5d)+(a+12d)=2a+17d=211\), while \(a_{22}=a+21d=256\). Doubling the second equation and subtracting the first gives \(25d=301\), so \(d=\frac{301}{25}\). Hence \(a=256-21d=\frac{79}{25}\). Therefore, \(a_{44}=a+43d=\frac{79}{25}+43\cdot\frac{301}{25}=\frac{13022}{25}\). The value \(539\) does not follow because the given data do not give \(d=14\). Exam tip: write \(a_n=a+(n-1)d\) carefully, especially when converting term indices.
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.
