The total number of elements in the first n rows is:
\[ S = 1 + 2 + 3 + \dots + T_n = \frac{n(n+1)}{2}. \]
To find the row containing 5310, solve:
\[ \frac{n(n+1)}{2} = 5310. \]
Start testing values:
\[ n = 100, \quad T_n = \frac{100 \cdot 101}{2} = 5050. \]
\[ n = 101, \quad T_n = \frac{101 \cdot 102}{2} = 5151. \]
\[ n = 102, \quad T_n = \frac{102 \cdot 103}{2} = 5253. \]
\[ n = 103, \quad T_n = \frac{103 \cdot 104}{2} = 5356. \]
Since 5310 lies between 5253 and 5356, it is in the 103rd row.
Final Answer: 103.
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The density of the copper ($^{64}Cu$) nucleus is greater than that of the carbon ($^{12}C$) nucleus.
Reason (R): The nucleus of mass number A has a radius proportional to $A^{1/3}$.
In the light of the above statements, choose the most appropriate answer from the options given below: