JEE Main 2025 — Sequences And Series Question with Solution

$$\begin{array}{ll}{1}^{\text{st }}\text{ A.P. }:1,6,11\dots & \Rightarrow T_n=S_n-4\\ 2^{\text{nd }}\text{ A.P.: }9,16,23\dots & \Rightarrow T_m=2+7m\end{array}$$

Let's find when they are equal for the first time:

$$\begin{array}{rl}& 5n-4=2+7m\\ & \Rightarrow 5n-7m=6\\ & \Rightarrow n=4,m=2\end{array}$$

$\Rightarrow 16$ is the first term, common difference will be

$$\mathrm{LCM}\left(d_1,d_2\right)=\mathrm{LCM}(5,7)=35$$

$\Rightarrow$ Common terms will be 16, 51, $86\dots$

The last term of $1^{\text{st }}$ A.P.

$$=T_{2025}=5\times 2025-4=10121$$

$$\begin{array}{rl}& \Rightarrow \text{ Common term must be less than that }\\ & \begin{array}{rl}& \Rightarrow 35n-19\\ & \Rightarrow 35n-19\le 10121\Rightarrow 35n\le 10140\\ & \Rightarrow n\le 289.7\\ & \Rightarrow n=289\\ & \Rightarrow \mathrm{in}n(A\cup B)=n(A)+n(B)-n(A\cap B)\\ & =2025+2025-289\\ & =3761\end{array}\end{array}$$