Let an arithmetic progression have 3n terms with a common difference of 2 . The sum of all the terms is 420 . If the sum of the terms at positions 1, 4, 7, , 3n-2 is 120 , then the value of n is equal to

Step-by-step solution

Let the arithmetic progression be a₁, a₂, a₃, , a_ 3n with common difference d = 2 . We can divide the terms into three groups based on their indices modulo 3 : S₁ = a₁ + a₄ + + a_ 3n-2 = 120 S₂ = a₂ + a₅ + + a_ 3n-1 S₃ = a₃ + a₆ + + a_ 3n Each group contains n terms. Notice the difference between corresponding terms of S₂ and S₁ : a₂ - a₁ = d, a₅ - a₄ = d, Thus, S₂ - S₁ = nd Similarly, S₃ - S₂ = nd S₃ = S₁ + 2nd Given d = 2 , we have: S₂ = S₁ + 2n = 120 + 2n S₃ = S₁ + 4n = 120 + 4n The sum of all terms is S₁ + S₂