The three numbers are arranged in the order of equal ratio sequence, the sum of which is 114. These three sequences are the first, the fourth and the 25th items of an arithmetic sequence in the order before. It is proved that B25 = 8b4-7b1

The three numbers are arranged in the order of equal ratio sequence, the sum of which is 114. These three sequences are the first, the fourth and the 25th items of an arithmetic sequence in the order before. It is proved that B25 = 8b4-7b1

Let me have a try... Let these three numbers A / Q, a, AQ form the equal ratio sequence a / Q + A + AQ = 114. From the problem a / Q, a, AQ are the first, fourth, and 25th terms of the equal ratio sequence {BN}, let the tolerance be Da = A / Q + 3daq = a + 21d → aq-a = 7 (A-A / Q) → Q-1 = 7 (1-1 / Q) = 7 (Q-1) / Q to get q = 1 or q = 71. Q = 1, 3A = 114, get a = 38, d =