Mathematics - Sequence & Series Question with Solution | TestHub

$a_n=a_1+(n-1)d$

$300=1+(n-1)d$
$\Rightarrow d=\frac{299}{(n-1)}=\frac{13\times 23}{(n-1)}=$ integer
so $n-1=\pm 13,\pm 23,\pm 299,\pm 1$
$\Rightarrow n=14,-12,24,-22,300,-298,2,0$
But $n\in [15, 50] \Rightarrow n=24 \Rightarrow d=13$
Hence,

$S_{n-4}=S_{20}=\frac{20}{2}\left[2\left(1\right)+(20-1)\left(13\right)\right]$

$\Rightarrow S_{n-4}=2490$
And,

$a_{n-4}=a_{20}=a_1+19d$

$=1+19\times 13$

$=248$