The sum of the three numbers is 114. If the three numbers are the first, fourth and 25th items of an arithmetic sequence in the previous order, then the sum of the numbers on each bit of the three numbers is?

The sum of the three numbers is 114. If the three numbers are the first, fourth and 25th items of an arithmetic sequence in the previous order, then the sum of the numbers on each bit of the three numbers is?

Let these three numbers be: a AQ AQ ^ 2, because these three sequences are the first, fourth, 25A + 3D = AQA + 24D = AQ ^ 2 of some arithmetic sequence in the previous order. By subtracting d from 8A + 24D = 8aqa + 24D = AQ ^ 2, we get: 7a = 8aq-aq ^ 2. By shifting term: Q ^ 2-8q + 7 = 0, we get q = 7 or q = 11. When q = 7a + AQ + AQ ^ 2 = 114, we get a = 2