QJEEM24Maf874: JEE Main 2024 Mathematics - Class 11 JEEM24Maf874 Solution: Let the positive integers be written in the form : If the k…

To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.

Understanding the Pattern

First row ( $k = 1$ ): Contains 1 number.

Second row ( $k = 2$ ): Contains 2 numbers.

Third row ( $k = 3$ ): Contains 3 numbers.

…

$k^\text{th}$ row: Contains $k$ numbers.

Total Numbers Up to the $k^\text{th}$ Row

The total number of integers up to the $k^\text{th}$ row is given by the sum of the first $k$ natural numbers:

$S(k) = 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2}$

Finding the Row Containing 5310

We need to find the smallest integer $k$ such that S(k-1) < 5310 \leq S(k) . This means:

\frac{(k-1)k}{2} < 5310 \leq \frac{k(k+1)}{2}

Step 1: Estimate $k$

Let's approximate $k$ by solving the inequality:

$\frac{k(k+1)}{2} \geq 5310$

Multiply both sides by 2:

$k(k+1) \geq 10620$

This is a quadratic inequality. We can approximate $k$ by taking the square root:

$k \approx \sqrt{10620} \approx 103$

Step 2: Calculate $S(k)$ for $k = 102$ and $k = 103$

For $k = 102$ :

$S(102) = \frac{102 \times 103}{2} = \frac{10506}{2} = 5253$

For $k = 103$ :

$S(103) = \frac{103 \times 104}{2} = \frac{10712}{2} = 5356$

Step 3: Determine the Correct Row

Since S(102) = 5253 < 5310 \leq 5356 = S(103) , the number $5310$ lies between $5254$ and $5356$ , inclusive. Therefore, it is in the $103^\text{rd}$ row.

Conclusion

The number $5310$ will be in the $103^\text{rd}$ row.