CAT 2019 Shift 2 Paper

The sequence is in arithmetic progression with first term $\left(\mathit{a}\right)=2\mathit{n}+1,$ common difference $\left(\mathit{d}\right)=2$ and last term $\left({\mathit{t}}_{\mathit{n}}\right)=2\mathit{n}+47$.
Let ' ${\mathit{m}}^{\prime }$ be the number of terms in this sequence.
The last term of A.P. is given by a+(n-1)d
$\Rightarrow \left(2\mathit{n}+1\right)+\left(\mathit{m}-1\right)\left(2\right)=2\mathit{n}+47$
$\Rightarrow \mathit{m}=24$
Also,
$=\frac{24}{2}\left[2\left(2\mathit{n}+1\right)+\left(24-1\right)\times 2\right]$
$=24\left(2\mathit{n}+1+23\right)=48\left(\mathit{n}+12\right)$
Therefore, $48\left(\mathit{n}+12\right)=5280\Rightarrow \mathit{n}=98$
Hence,