Mathematics · Sequence and Series

JEE Advanced 2018 — Paper 1 — Question 35

Let X be the set consisting of the first 2018 terms of the arithmetic progression $1,6,11, \ldots$ . , and Y be the set consisting of the first 2018 terms of arithmetic progression $9,16,23, \ldots .$ . . Then, the number of elements in the set $\mathrm{X} \cup \mathrm{Y}$ is $\_\_\_\_$ .

Answer: 3748

Numerical answer — enter this value.

Step-by-step solution

$n(X \cup Y)=n(X)+n(Y)-n(X \cap Y)$

$1,6,11, \ldots .2018$ term

\begin{aligned} & \mathrm{T}_{\mathrm{n}}=1+(\mathrm{n}-1) 5=5 \mathrm{n}-4 \\& \mathrm{U}_{\mathrm{K}}=9+(\mathrm{K}-1) 7=7 \mathrm{~K}+2 \end{aligned}

for common terms

\begin{aligned} & 5 \mathrm{n}-4=7 \mathrm{~K}+2 \\& \Rightarrow \quad \mathrm{n}=\frac{7 \mathrm{~K}+6}{5} \leq 2018, \mathrm{~K} \leq 1440 \\& \mathrm{~K}=2,7, \ldots \mathrm{r} \\& 2+(\mathrm{r}-1) 5 \leq 1440 \therefore \mathrm{r}_{\max }=288 \end{aligned}

$\therefore \quad$ number of common term $=288=\mathrm{n}(\mathrm{X} \cap \mathrm{Y})$

\mathrm{n}(\mathrm{X} \cup \mathrm{Y})=2018+2018-288=3748

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression