Total number of terms (n) = 37
∴ Middle term=(37+12)th term=19thterm
So, the three middle most terms are 18th, 19th and 20th terms.
By the given condition,
Sum of the three middle most terms = 225
a18+a19+a20=225
⇒(a+17d)+(a+18d)+(a+19d)=225
⇒3a+54d=225
⇒a+18d=75 ...(i)
And,
Sum of the last three terms = 429
⇒ a35+a36+a37=429
⇒ (a+34d)+(a+35d)+(a+36d)=429
⇒ 3a+105d=429
⇒ a+35d=143 ...(ii)
On subtracting Eq. (i) from Eq.(ii), we get,
17d = 68
⇒ d = 4
From Eq.(i),
a + 18(4) = 75
⇒ a=75−72
⇒ 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….