Explanation

(d)
Given that,
$a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
$\Rightarrow \ (a_{1} + a_{24} + \left( a_{5} + a_{20} \right) + \left( a_{10} + a_{15} \right) = 225$
$\Rightarrow \ 3\left( a_{1} + a_{24} \right) = 225$
$\Rightarrow \ a_{1} + a_{24} = 75$ …(i)
$(\because\$In an AP the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of the first and last term)
$\therefore\ \ a_{1} + a_{2} + ... + a_{24} = \frac{n}{2}\lbrack a + l)$
$= \frac{24}{2}(a_{1} + a_{24})$
$= 12 \times 75 = 900\$ [from Eq. (i)]