EPFO 2015 English Medium Solved Paper

Given, sequence, ${\mathit{a}}_1,{\mathit{a}}_2........,{\mathit{a}}_{24}$ is arithmetic progression.
Total terms = 24
We know that
${\mathit{T}}_{\mathit{n}}=\mathit{a}+\left(\mathit{n}-1\right)\mathit{d}$
According to the question,
${\mathit{a}}_1+{\mathit{a}}_5+{\mathit{a}}_{10}+{\mathit{a}}_{15}+{\mathit{a}}_{20}+{\mathit{a}}_{24}=225....\left(\mathit{i}\right)$
Let, first term = a
Then, ${\mathit{T}}_5={\mathit{a}}_5=\mathit{a}+4\mathit{d}$
${\mathit{T}}_{10}={\mathit{a}}_{10}=\mathit{a}+9\mathit{d}$
${\mathit{T}}_{15}={\mathit{a}}_{15}=\mathit{a}+14\mathit{d}$
${\mathit{T}}_{20}={\mathit{a}}_{20}=\mathit{a}+19\mathit{d}$
${\mathit{T}}_{24}={\mathit{a}}_{24}=\mathit{a}+23\mathit{d}$
From Eq. (i),
$\mathit{a}+\left(\mathit{a}+4\mathit{d}\right)+\left(\mathit{a}+9\mathit{d}\right)+\left(\mathit{a}+14\mathit{d}\right)+\left(\mathit{a}+190\right)+\left(\mathit{a}+23\mathit{d}\right)$
= $225$
⇒ $6\mathit{a}+69\mathit{d}=225$
$\Rightarrow 3\left(2\mathit{a}+23\mathit{d}\right)=225....\left(\mathit{i}\mathit{i}\right)$
Sum of arithmetic progression,
${\mathit{S}}_{\mathit{n}}=\frac{\mathit{n}}{2}\left[2\mathit{a}+\left(\mathit{n}-1\right)\mathit{d}\right]$
$=\frac{24}{2}\left[2\mathit{a}+\left(24-1\right)\mathit{d}\right]$
$=12\left[2\mathit{a}+23\mathit{d}\right]$
$=12\times \frac{225}{3}$ $$ $$ [form Eq.(ii)]
$=900$