Split 207 into three parts such that these

Let the three parts of the number 207 are (a – d), a and (a + d), which are in AP

Now, by given condition,

$\Rightarrow \quad$ Sum of these parts $=207$

$\Rightarrow \quad a-d+a+a+d=207$

$\Rightarrow \quad 3 a=207$

$a=69$

Given that, product of the two smaller parts $=4623$

$\Rightarrow \quad a(a-d)=4623$

$\Rightarrow \quad 69 \cdot(69-d)=4623$

$\Rightarrow \quad 69-d=67$

$\Rightarrow \quad \quad d=69-67=2$

So, $\quad$ first part $=a-d=69-2=67$,

second part $=a=69$

and $\quad$ third part $=a+d=69+2=71$,

Hence, required three parts are 67, 69, 71.