If the (21)st term of the AP (c-5,c+1,c+7,\ldots) is (139), what is the value of (c)?

Correct answer: 24

The difference between the first two terms is \((c+1)-(c-5)=6\), so \(d=6\) and the first term is \(a=c-5\). The 21st term is \(a_{21}=a+20d\). Hence, \(139=(c-5)+20\times6=c+115\), which gives \(c=24\). If \(c=26\), the 21st term would be 141, so it is not correct. Exam tip: for the \(n\)th term of an AP, use \(a+(n-1)d\).