The provided infinite series has a common ratio of \( \frac{1}{7} \) and can be represented as: \[ 7 = 5 + \frac{1}{7} (5 + \alpha) + \frac{1}{7^2} (5 + 2\alpha) + \cdots. \] This is a geometric series. The value of \( \alpha \) is determined by formulating the sum of the series and solving for \( \alpha \).
Final Answer: \( \alpha = \frac{6}{7} \).
\[ 5m \sum_{r=m}^{2m} T_r \text{ is equal to:} \]
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: