An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Let the three middle most terms of the AP be: a – d, a, a + d.

So,

(a – d) + a + (a + d) = 225

3a = 225

a = 75

Now, the AP is

a – 18d, …, a – 2d, a – d, a, a + d, a + 2d, …, a + 18d

Sum of last three terms:

(a + 18d) + (a + 17d) + (a + 16d) = 429

3a + 51d = 429

a + 17d = 143

75 + 17d = 143

d = 4

Now, first term = a – 18d = 75 – 18(4) = 3

Therefore, The AP is 3, 7, 11, … , 147.