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

Let the A.P. be a,a+d,a+2d,... .
Middle most term `=(37+1)/(2)`th term=19th term
`rArr` Three middle most terms `-a_(18),a_(19) and a_(20)`
Given that, `a_(18)+a_(19)+a_(20)=225`
`rArr (a+17d)+(a+18d)+(a+19d)=225`
`rArr 3a+54d=225`
`rArr a+18d=75 `
and `a_(35)+a_(36)+a_(37)=429" " ...(1)`
`rArr (a+34d)+(a+35d)+(a+36d)=429`
`rArr 3a+105d=429 rArr a+35d=143`
`rArr 75-18+35d=143 " "` [from (1)]
`rArr 17d=68 rArr d=4`
Put d=4 in equation (1), we get
`a+18(4)=75 rArr a=3`
`:. A.P. =3,7,11,15, ...`