JEE 2024 — Mathematics PYQ

First A.P. 4, 9, 14, 19, ..., 48 to $25^{th}$ term

Here, $a=4$, $d=5$

So, $a_{25}=4+24(5)=124$

Second A.P. 3, 6, 9, 12, ..., 48 to $7^{th}$ term

Here, $a=3$, $d=3$

So, $a_7=3+36(3)=111$

As, LCM (5, 3) = 15

So, we have, $9+(x-1)\le 111$

$\Rightarrow 9+15n-15\le 111$

$\Rightarrow 15n-6\le 111$

$\Rightarrow 15n\le 117$

$\Rightarrow n\le 7.4$. So, $n=7$