Q.5. Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference

Q: What is the correct answer for JEE Main Mathematics Q5?

The correct answer is B. Sol. For set A: First term $a_1 = 1$, common difference $d_1 = 5$. Terms are $1, 6, 11, 16, \dots$ $n_A = 101$. Last term of A: $1 + (101-1)5 = 1 + 100 \times 5 = 501$. So, $A = \{1, 6, 11, \dots, 501\}$. For set B: First term $a_2 = 9$, common difference $d_2 = 7$. Terms are $9, 16, 23, \dots$ $n_B = 71$. Last term of B: $9 + (71-1)7 = 9 + 70 \times 7 = 9 + 490 = 499$. So, $B = \{9, 16, 23, \dot

Q: Q.5. 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…

The correct answer is B. Sol. For set A: First term $a_1 = 1$, common difference $d_1 = 5$. Terms are $1, 6, 11, 16, \dots$ $n_A = 101$. Last term of A: $1 + (101-1)5 = 1 + 100 \times 5 = 501$. So, $A = \{1, 6, 11, \dots, 501\}$. For set B: First term $a_2 = 9$, common difference $d_2 = 7$. Terms are $9, 16, 23, \dots$ $n_B = 71$. Last term of B: $9 + (71-1)7 = 9 + 70 \times 7 = 9 + 490 = 499$. So, $B = \{9, 16, 23, \dot

Question

Q.5. 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 , which are divisible by 3, is :

FirstInTest · Accepted Answer

Sol. For set A:
First term $a_1 = 1$, common difference $d_1 = 5$.
Terms are $1, 6, 11, 16, \dots$
$n_A = 101$.
Last term of A: $1 + (101-1)5 = 1 + 100 	imes 5 = 501$.
So, $A = \{1, 6, 11, \dots, 501\}$.

For set B:
First term $a_2 = 9$, common difference $d_2 = 7$.
Terms are $9, 16, 23, \dots$
$n_B = 71$.
Last term of B: $9 + (71-1)7 = 9 + 70 	imes 7 = 9 + 490 = 499$.
So, $B = \{9, 16, 23, \dots, 499\}$.

For $A \cap B$, the common difference $D = 	ext{LCM}(d_1, d_2) = 	ext{LCM}(5, 7) = 35$.
The first common term is 16.
Let the terms in $A \cap B$ be $T_k = 16 + (k-1)35$.
We need $T_k \le 499$ (since the maximum term in B is 499, and in A is 501).
$16 + (k-1)35 \le 499$
$(k-1)35 \le 483$
$k-1 \le \frac{483}{35} \approx 13.8$
$k-1 \le 13 \Rightarrow k \le 14$.
So there are 14 terms in $A \cap B$. These terms are:
$16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471$.

Now we need to find the terms in $A \cap B$ that are divisible by 3.
We check each term:
$16 \div 3$ (remainder 1)
$51 \div 3 = 17$ (divisible)
$86 \div 3$ (remainder 2)
$121 \div 3$ (remainder 1)
$156 \div 3 = 52$ (divisible)
$191 \div 3$ (remainder 2)
$226 \div 3$ (remainder 1)
$261 \div 3 = 87$ (divisible)
$296 \div 3$ (remainder 2)
$331 \div 3$ (remainder 1)
$366 \div 3 = 122$ (divisible)
$401 \div 3$ (remainder 2)
$436 \div 3$ (remainder 1)
$471 \div 3 = 157$ (divisible)

The terms divisible by 3 are $\{51, 156, 261, 366, 471\}$.
There are 5 such terms.

Answer

4

Answer

5

Answer

6

Answer

7

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