Let A₁, A₂, A₃, , A₃₉ be 39 arithmetic means between the numbers 59 and 159 . Then the mean of A₂₅, A₂₈, A₃₁ and A₃₆ is equal to :

Step-by-step solution

The sequence 59, A₁, A₂, , A₃₉, 159 forms an arithmetic progression with 41 terms. The common difference d is given by d = 159 - 59 39 + 1 = 100 40 = 2.5 The k -th arithmetic mean is A_k = 59 + k d The required mean is A₂₅ + A₂₈ + A₃₁ + A₃₆ 4 A₂₅ + A₂₈ + A₃₁ + A₃₆ 4 = (59 + 25d) + (59 + 28d) + (59 + 31d) + (59 + 36d) 4 = 4 59 + 120d 4 = 59 + 30d Substituting d = 2.5 : = 59 + 30(2.5) = 59 + 75 = 134 Answer: 134