Step 1: Given that the difference between consecutive terms \( a_{n+1} - a_n \) bears a constant ratio with the sum of consecutive terms \( a_n + a_{n+1} \), we express this as:
\[ \frac{a_{n+1} - a_n}{a_{n+1} + a_n} = k \]
where \( k \) is a constant ratio.
Step 2: Rearranging this equation:
\[ a_{n+1} - a_n = k(a_{n+1} + a_n) \]
Step 3: Simplifying:
\[ a_{n+1} - a_n = k a_{n+1} + k a_n \] \[ a_{n+1} - k a_{n+1} = a_n + k a_n \] \[ a_{n+1}(1 - k) = a_n(1 + k) \]
Step 4: Solving for \( a_{n+1} \):
\[ a_{n+1} = \frac{1 + k}{1 - k} a_n \]
This is a characteristic property of a series where the ratio between consecutive terms is constant. This behavior corresponds to a harmonic progression (H.P.), because the general term \( a_n \) satisfies the relation of terms being inversely proportional to an arithmetic progression.
If the sum of the first 10 terms of the series \[ \frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \ldots \] is \(\frac{m}{n}\), where \(\gcd(m, n) = 1\), then \(m + n\) is equal to _____.
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :

A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: