If the sum of 12th and 22nd terms of an AP is 100, then the sum of the first 33 terms of an AP is

Step-by-step solution

Let the first term of the arithmetic progression be a and the common difference be d . The n -th term of an AP is given by T_n = a + (n-1)d . The 12th term is T₁₂ = a + 11d and the 22nd term is T₂₂ = a + 21d . Given T₁₂ + T₂₂ = 100 , we have (a + 11d) + (a + 21d) = 100 , which simplifies to 2a + 32d = 100 . Dividing by 2, we get a + 16d = 50 . The sum of the first n terms of an AP is given by S_n = n 2 [2a + (n-1)d] . For n = 33 , the sum is S₃₃ = 33 2 [2a + (33-1)d] = 33 2 [2a + 32d] . Substituting 2a + 32d = 100