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

Since, total number of terms (n) `= 37 " " ` [odd]
` :.` Middle term `=((37+1)/(2))`th term = 19 th term
So, the three middle most terms = 18th, 19th and 20th.
By given condition,
Sum of the three middle most terms = 225
`a_(18)+a_(19)+a_(20) = 225`
`implies (a+17d)+(a+18d)+(a+19d) =225`
`implies 3a+54d=225`
` implies a +18d=75 " "` ...(i)
and sum of the last three terms = 429
`a_(35)+a_(36)+a_(37) = 429`
`implies (a+34d)+(a+35d)+(a+36d) =429`
`implies 3a+105d=429`
` implies a +35d=143 " "` ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
`17d=68`
`implies d=4`
From Eq. (i), `a+18(4)=75`
`implies a=75-72`
`implies a=3`
` :. ` Required AP is `a, a+d, a+2d, a+3d, ...`
i.e., `3, 3+4, 3+2(4), 3+3(4), ...`
i.e., `3, 7, 3+8, 3+12, ...`
i.e., `3, 7, 11, 15, ...`