In an AP, (a_{18}=0) and (a_{42}=-192). What is (a_1)?
Advertisement
Answer and explanation
Correct answer: \(136\)
For an AP, \(a_{42}-a_{18}=(42-18)d\). Thus, \(-192-0=24d\), so \(d=-8\). Using \(a_{18}=a_1+17d\), we get \(0=a_1+17(-8)\), hence \(a_1=136\). If \(128\) were used, the 18th term would not be 0. Exam tip: Find \(d\) first by subtracting the two given terms.
Frequently asked questions
What is the correct answer to this question?
\(136\)
Why is this the correct answer?
For an AP, \(a_{42}-a_{18}=(42-18)d\). Thus, \(-192-0=24d\), so \(d=-8\). Using \(a_{18}=a_1+17d\), we get \(0=a_1+17(-8)\), hence \(a_1=136\). If \(128\) were used, the 18th term would not be 0. Exam tip: Find \(d\) first by subtracting the two given terms.
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.
Advertisement