If the (5)th term of an AP is (x+7) and the (12)th term is (x+42), what is the (20)th term in terms of (x)?

Correct answer: \(x+82\)

There are 7 positions between the 5th and 12th terms. Thus, \(a_{12}-a_5=7d\), so \((x+42)-(x+7)=35=7d\), giving \(d=5\). From the 12th term to the 20th term, there are 8 positions; hence \(a_{20}=a_{12}+8d=x+42+8(5)=x+82\). The option \(x+77\) results from incorrectly using a gap of only 7 positions. Exam tip: the difference between term numbers gives the multiplier of \(d\).