JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Sequences and Series

Let

X

be the set consisting of the first 2018 terms of the arithmetic progression

1, 6, 11, \dots

, and

Y

be the set consisting of the first 2018 terms of the arithmetic progression

9, 16, 23, \dots

. Then, the number of elements in the set

X \cup Y

is ____.

Answer: 3748

Solution

For

X

a = 1, d = 5

. The 2018th term is

1 + 2017 \times 5 = 10086

.

For

Y

a = 9, d = 7

. The 2018th term is

9 + 2017 \times 7 = 14128

.

Common terms: Both sequences can be written as

1 + 5 k

and

9 + 7 m

. Setting equal:

1 + 5 k = 9 + 7 m

gives

5 k - 7 m = 8

. The general solution is

k = 16 + 7 n

, giving terms

16, 51, 86, \dots

with common difference 35.

The largest common term

\leq 10086

16 + 35 (n - 1) \leq 10086

gives

n \leq 288.7

, so 288 common terms.

By inclusion-exclusion:

∣ X \cup Y ∣ = 2018 + 2018 - 288 = 3748

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer