Maths Q5 – JEE-MAIN 2026

Home Maths Sequences and SeriesQuestion #389

Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in A

\cap

B, which are divisible by 3, is:

Clue 1 of 3: Arithmetic Progressions and Set Intersection

First, determine the general terms and ranges for sets A and B. Then, find the general form of elements common to both sets A and B.

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● Video Walkthrough

Step 1: Define the sets A and B and their general terms✦ Active

Set A has first term $a_{1} = 1$ , common difference $d_{1} = 5$ , and $n_{1} = 101$ terms. Its general term is $A_{k} = a_{1} + (k - 1) d_{1} = 1 + (k - 1) 5 = 5 k - 4$ , for $1 \leq k \leq 101$ . The terms of A range from $A_{1} = 1$ to $A_{101} = 1 + 100 \times 5 = 501$ .

Set B has first term $a_{2} = 9$ , common difference $d_{2} = 7$ , and $n_{2} = 71$ terms. Its general term is $B_{m} = a_{2} + (m - 1) d_{2} = 9 + (m - 1) 7 = 7 m + 2$ , for $1 \leq m \leq 71$ . The terms of B range from $B_{1} = 9$ to $B_{71} = 9 + 70 \times 7 = 499$ .

Step 2: Find the general form of elements in

A \cap B

and their range○ Expand

For an element to be in $A \cap B$ , we must have $A_{k} = B_{m}$ . This leads to the equation $5 k - 4 = 7 m + 2$ , which simplifies to $5 k - 7 m = 6$ .

A particular solution to this linear Diophantine equation is $(k, m) = (4, 2)$ , since $5 (4) - 7 (2) = 20 - 14 = 6$ . The general solution for $k$ and $m$ is $k = 4 + 7 t$ and $m = 2 + 5 t$ , where $t$ is an integer.

Substituting $k$ (or $m$ ) back into the general term, the common terms are $X_{t} = 5 (4 + 7 t) - 4 = 35 t + 16$ .

Now, we determine the range for $t$ based on the constraints for $k$ and $m$ :

1 \leq 4 + 7 t \leq 101 ⟹ - 3 \leq 7 t \leq 97 ⟹ - \frac{3}{7} \leq t \leq \frac{97}{7} ⟹ - 0.42 \leq t \leq 13.85

1 \leq 2 + 5 t \leq 71 ⟹ - 1 \leq 5 t \leq 69 ⟹ - \frac{1}{5} \leq t \leq \frac{69}{5} ⟹ - 0.2 \leq t \leq 13.8

Combining these conditions, $t$ must be an integer in the range $0 \leq t \leq 13$ . Thus, $t \in {0, 1, 2, \dots, 13}$ . There are $13 - 0 + 1 = 14$ common terms in $A \cap B$ .

💡 Teacher's Secret Hint

Ensure to find the intersection of the ranges for 't' derived from both sets A and B.

Step 3: Find the number of common elements divisible by 3○ Expand

We need to find terms $X_{t} = 35 t + 16$ that are divisible by 3. This can be expressed using modulo arithmetic:

35 t + 16 \equiv 0 (\mod 3)

Since $35 \equiv 2 (\mod 3)$ and $16 \equiv 1 (\mod 3)$ , the congruence becomes:

2 t + 1 \equiv 0 (\mod 3)

2 t \equiv - 1 (\mod 3)

2 t \equiv 2 (\mod 3)

Dividing by 2 (which is coprime to 3), we get:

t \equiv 1 (\mod 3)

We need to find the values of $t$ in the set ${0, 1, 2, \dots, 13}$ that satisfy $t \equiv 1 (\mod 3)$ . These values are $t = 1, 4, 7, 10, 13$ .

There are 5 such values of $t$ . Therefore, there are 5 elements in $A \cap B$ that are divisible by 3.

💡 Teacher's Secret Hint

Remember to check the divisibility condition for the common terms, not the original AP terms directly.

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Arithmetic Progressions and Set Intersection

🧠 Full Solution Path