Question

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:

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(7\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The intersection of two arithmetic progressions (A.P.s) is also an A.P. whose common difference is the least common multiple (LCM) of the common differences of the two original series.
We need to find common terms that satisfy the range constraints of both sets and then count those that are divisible by 3.
Step 2: Key Formula or Approach:
1. General term of A.P.: \(T_n = a + (n-1)d\).
2. Common difference of intersection (\(d_c\)): \(LCM(d_1, d_2)\).
3. Range constraint: \(T_n \leq \min(\text{max value of A, max value of B})\).
Step 3: Detailed Explanation:
For set A: First term \(a_1 = 1\), difference \(d_1 = 5\), number of terms \(n_1 = 101\).
Max term of A = \(1 + (101-1)5 = 501\).
For set B: First term \(a_2 = 9\), difference \(d_2 = 7\), number of terms \(n_2 = 71\).
Max term of B = \(9 + (71-1)7 = 499\).
Common terms (\(A \cap B\)) must be \(\leq 499\).
Let's find the first common term:
Terms of A: 1, 6, 11, 16, 21, ...
Terms of B: 9, 16, 23, ...
First common term \(a = 16\).
Common difference \(d_c = LCM(5, 7) = 35\).
Common terms are of form: \(16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471, 506 \dots\)
Within range \(\leq 499\), terms are: 16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471.
Now filter for divisibility by 3:
- 51 (sum of digits \(5+1=6\)): Yes
- 156 (sum of digits \(1+5+6=12\)): Yes
- 261 (sum of digits \(2+6+1=9\)): Yes
- 366 (sum of digits \(3+6+6=15\)): Yes
- 471 (sum of digits \(4+7+1=12\)): Yes
Total count = 5.
Step 4: Final Answer:
The number of elements in \(A \cap B\) divisible by 3 is 5.