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.

Explanation

Total number of terms (n)=37, which is odd
`:.` Middle term `=(37+1)/(2)`th term = 19th term
So, 3 middle most terms are 18th, 19th and 20th
`:. T_(18)+T_(19)+T_(20)=225`
`rArr a+17d+a+18d+a+19d=225`
`rArr 3a+54d=225 rArr a+18d=75 " " ...(1)`
Also, sum of last 3 term =429
`T_(35)+T_(36)+T_(37)=429`
`rArr a+34d+a+35d+a+36d=429`
`rArr 3a+105d=429 rArr a+35d=143 " " ...(2)`
Solving, (1) and (2), we get d=4 and a=3
`:.` Required A.P. is a, a+d, a+2d, a+3d,...
i.e., 3,3+4,3+2(4),3+3(4)
i.e., 3,7,11,15,...