JEE Main 2024 — 6 April, Shift 1 — Question 6

$\mathrm{n}(3) \Rightarrow$ multiple of 3

102, 105, 108………, 699

$\mathrm{T}_{\mathrm{n}}=699=102+(\mathrm{n}-1)(3)$

$\mathrm{n}=200$

$\mathrm{n}(3)=200$

$\because \mathrm{n}(4) \Rightarrow$ multiple of 4

$100,104,108, \ldots ., 700$

$\mathrm{T}_{\mathrm{n}}=700=100+(\mathrm{n}-1)(4)$

$\mathrm{n}=151$

$\mathrm{n}(4)=151$

$\mathrm{n}(3 \cap 4) \Rightarrow$ multiple of $3 \& 4$ both $

$108,120,132, \ldots ., 696$

$\mathrm{T}_{\mathrm{n}}=696=108+(\mathrm{n}-1)(12)$

$\mathrm{n}=50$

$\begin{aligned} & \mathrm{n}(3 \cap 4)=50 & \end{aligned}$

$\mathrm{n}(3 \cup 4)=\mathrm{n}(3)+\mathrm{n}(4)-\mathrm{n}(3 \cap 4) =200+151-50 =301$

$\mathrm{n}(\overline{3 \cup 4})=$ Total $-\mathrm{n}(3 \cup 4)=$ neither a multiple of 3 nor a multiple of 4

$=601-301=300$