Given:Sum of the first 9 terms of an AP = sum of the first 11 termsFormula used:Sum of n terms of AP:Sn = \(\frac{n}{2}[2a+ (n-1)d]\)Where a and d are the first terms and common differences respectively.Calculation:Let, a and d be first terms and common differences respectivelyAccording to the question, S9 = S11By using the formula of the sum of AP\(? \frac{9}{2}[2a+8d]=\frac{11}{2}[2a+10d]\)? 18a + 72d = 22a + 110d? 4a = - 38d? a = \(\frac{-19d}{2}\) -----(1)Now, sum of the first 20 terms\({{S}_{20}}=\frac{20}{2}\left[ 2a+19d \right]\)From equation (1)S20 = 10 [2 × (\(\frac{-19d}{2}\)) + 19d]? S20 = 10[19d + 19d] = 0? The sum of the first 20 terms is 0.