Concept Used:1). The nth term of AP is given by Tn = a + (n “ 1) × d2). Sum of n terms in AP is given by Sn = \(\frac{n}{2}\)[2a + (n ? 1) × d]3). Sum of n terms in a finite AP with the last term as l is given by S = \(\frac{n}{2}\)(a + l),Where a = First term, n = nth term, d = common difference, l = last term.4). \(\displaystyle Mean = \frac{Sum \ of\ all \ the\ observations}{Total\ number \ of \ observations}\)Calculation:Given:n = 51, a = 4, d = 3.Sum of n terms in AP is given by Sn = \(\frac{n}{2}\)[2a + (n ? 1) × d]? Sn = \(\frac{n}{2}\)[2a + (n ? 1) × d]? S51 = \(\frac{51}{2}\)[2 × 4 + (51 ? 1) × 3]? S51 = \(\frac{51}{2}\)[8 + 50 × 3]? S51 = \(\frac{51}{2}\) × 158? S51 = 51 × 79\(\displaystyle Mean = \frac{Sum \ of\ all \ the\ observations}{Total\ number \ of \ observations}\)? Mean = \(\displaystyle \frac{51 \times 79}{51}\)? Mean = 79? The mean of the marks is 79.