(a) Start by expressing \(\frac{1}{(ar+1)(ar+a+1)}\) using partial fractions:

\(\frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{ar+1} - \frac{1}{ar+a+1} \right)\)

Now, sum from \(r=1\) to \(n\):

\(\sum_{r=1}^{n} \frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{a+1} + \frac{1}{2a+1} + \frac{1}{3a+1} + \ldots + \frac{1}{an+a+1} \right)\)

Using the method of differences, this simplifies to:

\(\frac{1}{a} \left( \frac{1}{a+1} - \frac{1}{an+a+1} \right)\)

(b) For the infinite series:

\(\sum_{r=1}^{\infty} \frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{a+1} \right) = \frac{1}{6}\)

This leads to:

\(\frac{1}{a^2 + a} = \frac{1}{6}\)

Solving \(a^2 + a - 6 = 0\) gives \(a = 2\).