The first term of an arithmetic progression is (15), the last term is (51), and the number of terms is (13). Find the sum.

Correct answer: (429)

The sum of n terms of an arithmetic progression can be found from the first term, last term, and number of terms using \(S_n=\frac{n}{2}(a+l)\). This works because the terms can be paired from the two ends, and every pair has the same total. Equivalently, the sum equals the number of terms multiplied by the average of the first and last terms.

Here \(n=13\), the first term is \(a=15\), and the last term is \(l=51\). Therefore, \(S_{13}=\frac{13}{2}(15+51)=\frac{13}{2}\times66=13\times33=429\). The average term is \((15+51)/2=33\), and 13 times 33 also gives 429. Hence option B, 429, is correct. The number of terms must be 13, not the number of gaps between them.