In an AP, (a_4+a_{16}=86) and (a_{10}=43). Find (a_{25}) when the common difference is (5).
Answer and explanation
Correct answer: 118
Use the AP relation from a known term: \(a_n=a_r+(n-r)d\). Thus, \(a_{25}=a_{10}+(25-10)d=43+15\times5=118\). Hence, option B is correct. The condition \(a_4+a_{16}=86\) confirms \(2a_{10}=86\), since the 4th and 16th terms are equally placed around the 10th term; it is not needed separately to find \(a_{25}\). Exam tip: use the index difference \(25-10=15\), not \(25-1\).
Frequently asked questions
What is the correct answer to this question?
118
Why is this the correct answer?
Use the AP relation from a known term: \(a_n=a_r+(n-r)d\). Thus, \(a_{25}=a_{10}+(25-10)d=43+15\times5=118\). Hence, option B is correct. The condition \(a_4+a_{16}=86\) confirms \(2a_{10}=86\), since the 4th and 16th terms are equally placed around the 10th term; it is not needed separately to find \(a_{25}\). Exam tip: use the index difference \(25-10=15\), not \(25-1\).
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.
