Let A=1,6,11,16, and B=9,16,23,30, be the sets consisting of the first 2025 terms of two a…

Question

Let $A=\{1,6,11,16, \ldots\}$ and $B=\{9,16,23,30, \ldots\}$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n(A \cup B)$ is

Accepted Answer

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

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

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

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

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

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

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

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

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

Solutions · 1