Given:5th term of an AP is \(\frac{1}{10}\) and its 10th term is \(\frac{1}{5},\)Formula used:nth term of AP:Tn = a + (n - 1)dSum of n terms of AP:Sn = \(\frac{n}{2}\)[2a + (n - 1)d]Where a and d are the first term and common differences of AP.Calculation:According to the question,T5 = \(\frac{1}{10}\)Using the formula of nth term of AP? T5 = a + 4d ? a + 4d = \(\frac{1}{10}\) ------(1)Similarly, we havea + 9d = \(\frac{1}{5}\) ------(2)Doing equation (2) - (1)? (a + 9d) - (a + 4d) = \(\frac{1}{5}\) - \(\frac{1}{10}\)? 5d = \(\frac{1}{10}\)? d = \(\frac{1}{50}\)Put this value in equation (1)a + \(4×\frac{1}{50}\) = \(\frac{1}{10}\)? a = \(\frac{1}{10}- \frac{2}{25} = \frac{1}{50}\)Now using the formula, Sn = \(\frac{n}{2}\)[2a + (n - 1)d]Sum of 50 terms of AP = S50 = \(\frac{50}{2}\)[2×\(\frac{1}{50}\) + (50 - 1)\(\frac{1}{50}\)]? S50 = 25\((\frac{2}{50}+ \frac{49}{50} )= \frac{51}{2}\) = 25.5? The sum of the first 50 terms is 25.5.