Let the positive integers be written in the f | vedclass

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(A) The total number of elements up to the $n^{	ext{th}}$ row is given by the sum of the first $n$ natural numbers: $S_n = 1 + 2 + 3 + \dots + n = \f

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$103$

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(A) The total number of elements up to the $n^{	ext{th}}$ row is given by the sum of the first $n$ natural numbers: $S_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$.The last number in the $n^{	ext{th}}$ row is $S_n = \frac{n(n+1)}{2}$.We want to find the row $n$ such that the number $5310$ lies in it. This means the number $5310$ must satisfy $S_{n-1} $\frac{(n-1)n}{2} $(n-1)n For $n^2 \approx 10620$,we have $n \approx \sqrt{10620} \approx 103.05$.Let's test $n = 103$:$S_{103} = \frac{103 	imes 104}{2} = 103 	imes 52 = 5356$.$S_{102} = \frac{102 	imes 103}{2} = 51 	imes 103 = 5253$.Since $5253 < 5310 \le 5356$,the number $5310$ lies in the $103^{	ext{rd}}$ row.

Let the positive integers be written in the form:
$1$
$2$ $3$
$4$ $5$ $6$
$7$ $8$ $9$ $10$
If the $k^{\text{th}}$ row contains exactly $k$ numbers for every natural number $k$,then the row in which the number $5310$ will be,is:

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Let the positive integers be written in the form:$1$$2$ $3$$4$ $5$ $6$$7$ $8$ $9$ $10$If the $k^{\text{th}}$ row contains exactly $k$ numbers for every natural number $k$,then the row in which the number $5310$ will be,is: