Concept:1). The nth term of AP is given by Tn = a + (n “ 1) × dWhere a = First term, n = nth term, d = common difference.2). Deviation = ? (X “ µ)Here, ? represents the addition of valuesX represents each value in the data setµ represents the mean of the data setCalculation:The nth term of AP is given by Tn = a + (n “ 1) × dTn = 4 + (51 “ 1) × 3Tn = 154 ----(i)So, A.P must be 4, 7, 10, ......... 151, 154 ----(ii)Now, Sum of n terms in AP is given by Sn = \(\frac{n}{2}\)[2a + (n ? 1) × d]? Sn = \(\frac{51}{2}\)[2 × 4 + (51 ? 1) × 3]? Sn = 4029\(\displaystyle Mean = \frac{Sum \ of\ all \ the\ observations}{Total\ number \ of \ observations}\)? Mean = \(\displaystyle \frac{4029}{51}\)? Mean = 79 ----(iii)Now, Sum of the deviations of the marks measured from median is given byFrom (i), (ii) and (iii), we can write(4 - 79) + (7 - 79) + (10 - 79) + .........+ (151 - 79) + (154 - 79)? (-75) + (-72) + (-69) + .........+ (72) + (75) = 0? The sum of the deviations measured from the median is 0.