In an AP (a_{10}+a_{30}=500) and (a_{18}+a_{38}=820). What is (a_{58})?
Answer and explanation
Correct answer: 1010
Let the first term be \(a\) and common difference be \(d\). Then \(a_{10}+a_{30}=2a+38d=500\) and \(a_{18}+a_{38}=2a+54d=820\). Subtracting gives \(16d=320\), so \(d=20\). From \(2a+38(20)=500\), we get \(a=-130\). Hence, \(a_{58}=a+57d=-130+57(20)=1010\). The nearby option 1030 can result from incorrectly using \(58d\); remember that \(a_{58}=a+57d\). Exam tip: use \(a_n=a+(n-1)d\).
Frequently asked questions
What is the correct answer to this question?
1010
Why is this the correct answer?
Let the first term be \(a\) and common difference be \(d\). Then \(a_{10}+a_{30}=2a+38d=500\) and \(a_{18}+a_{38}=2a+54d=820\). Subtracting gives \(16d=320\), so \(d=20\). From \(2a+38(20)=500\), we get \(a=-130\). Hence, \(a_{58}=a+57d=-130+57(20)=1010\). The nearby option 1030 can result from incorrectly using \(58d\); remember that \(a_{58}=a+57d\). Exam tip: use \(a_n=a+(n-1)d\).
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.
