An AP consists of 37 terms.

Solution:

Since, total number of terms                                                                                                                 [Odd]

$\therefore$ Middle term $=\left(\frac{37+1}{2}\right)$ th term $=19$ th term

So, the three middle most terms $=18$ th 19 th and 20 th. By given condition,

Sum of the three middle most terms $=225$

$a_{18}+a_{19}+a_{20}=225$

$\Rightarrow \quad(a+17 d)+(a+18 d)+(a+19 d)=225$

$\Rightarrow \quad 3 a+54 d=225$

$\Rightarrow \quad a+18 d=75$ ...(i)

and sum of the last three terms $=429$

$\Rightarrow \quad a_{35}+a_{36}+a_{37}=429$

$\Rightarrow \quad(a+34 d)+(a+35 d)+(a+36 d)=429$

$\Rightarrow \quad 3 a+105 d=429$

$\Rightarrow \quad a+35 d=143 \quad \ldots$ (ii)

On subtracting Eq. (i) from Eq. (ii), we get

$17 d=68$

$\Rightarrow \quad d=4$

From Eq. (i), $a+18(4)=75$

$\Rightarrow \quad a=75-72$

$\Rightarrow \quad a=3$

$\therefore$ Required AP is $a_{1} a+d, a+2, a+3 d, \ldots$

i.e. $\quad 3.3+43+2(4), 3+3(4), \ldots$

i.e., $\quad 3,7,3+8,3+12, \ldots$

i.e., $3,7,11,15, \ldots$