We know that the sum of terms of AP equidistant from the beginning and end is always same and it is always equal to the sum of first and last terms.
$\Rightarrow \, a_{1}+a_{24}=a_{5}+a_{20}=a_{10}+a_{15}$
$\because \,a_{1}+a_{5}+a_{10}+a_{15}+a_{20}+a_{24}=225$
$\therefore \, 3\left(a_{1}+a_{24}\right)=225$
$ \Rightarrow \, a_{1}+a_{24}=75$
$\therefore \, S_{24}=\frac{24}{2}\left(a_{1}+a_{24}\right)$
$\left[\because S_{n}=\frac{n}{2}\left(a_{1}+a_{n}\right)\right]$
$=12(75)=900=9 \times 10^{2}$