In an AP, (a_5+a_{15}=100) and (a_9+a_{19}=164). What is (a_{27})?
Answer and explanation
Correct answer: 186
Let the first term be \(a\) and the common difference be \(d\). Then \(a_5+a_{15}=2a+18d=100\) and \(a_9+a_{19}=2a+26d=164\). Subtracting the first equation from the second gives \(8d=64\), so \(d=8\). Using \(2a+18\times8=100\), we get \(a=-22\). Hence, \(a_{27}=a+26d=-22+26\times8=186\). The value 210 does not follow from the correct common difference. Exam tip: subtract equations formed from sums of AP terms to find \(d\) quickly.
Frequently asked questions
What is the correct answer to this question?
186
Why is this the correct answer?
Let the first term be \(a\) and the common difference be \(d\). Then \(a_5+a_{15}=2a+18d=100\) and \(a_9+a_{19}=2a+26d=164\). Subtracting the first equation from the second gives \(8d=64\), so \(d=8\). Using \(2a+18\times8=100\), we get \(a=-22\). Hence, \(a_{27}=a+26d=-22+26\times8=186\). The value 210 does not follow from the correct common difference. Exam tip: subtract equations formed from sums of AP terms to find \(d\) quickly.
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.
