JEE 2023 — Mathematics PYQ

Method of difference:
$S_n=5+8+14+23+\cdots +a_n$
$S_n=0+5+8+14+\cdots +a_{n-1}+a_n$
Subtracting: $0=5+3+6+9+\cdots -a_n$
$\Rightarrow a_n=5+[3+6+9+\dots (n-1)\text{ terms}]$
$=5+\left[\frac{(n-1)}{2}(6+(n-2)3)\right]$

$a_n=5+\left[\frac{(n-1)}{2}(6+(n-2)3)\right]=\frac{3n^2-3n+10}{2}$

So, $a_{40}=\frac{3(40{)}^2-3(40)+10}{2}=2345$

$S_{30}=\frac{3{\sum }_{n=1}^{30}{n}^2-3{\sum }_{n=1}^{30}n+10{\sum }_{n=1}^{30}1}{2}=13635$

$\therefore S_{30}-a_{40}=13635-2345=11290$