If the sum of the first 11 terms of an arithmetic progression is equal to the sum of its first 19 terms, then the sum of the first 30 terms will be-

Explanation

The sum of n terms of an arithmetic progression = $$\frac{n}{2} [2a + (n - 1)d]$$ where a = first, d = common difference n = number of terms ∴ 11 Sum of terms = $$\frac{11}{2} [2a + 10d]$$ = 11a + 55d ....(i) Sum of 19 terms = $$\frac{19}{2} [2a + 18d]$$ = 19a + 171d .....(ii) As per the question, 11a + 55d = 19a + 171d 8a = 55d – 171d 8a = – 116d a = -$$\frac{116}{8}$$d = -$$\frac{29}{2}$$d Now, the sum of 30 terms = $$\frac{30}{2} (2a + 29d)$$ = 15[2 × (-$$\frac{29}{2}$$d) + 29d] = 15[-29d + 29d] = 15 × 0 = 0