2011 AJC P1 Q12

The sum of the first \(n\) terms of a series, \({{S}_{n}}\), is given by

\({{S}_{n}}=\frac{1}{a}\left[ 1-{{(a-1)}^{n}} \right]\), where \(a\) is a constant and \(a\ne 1\), \(n\in {{\mathbb{Z}}^{+}}\)

Obtain an expression for the \({{n}^{\mathrm{th}}}\) term of the series, \({{T}_{n}}\), and prove that \({{T}_{n}}\) is a geometric series.[3]

If the sequence \({{T}_{n}}\) is now grouped as follows: \(({{T}_{1}}),\ ({{T}_{2}},\ {{T}_{3}},\ {{T}_{4}}),\ ({{T}_{5}},\ {{T}_{6}},\ {{T}_{7}},\ {{T}_{8}},\ {{T}_{9}}),\ \ldots \) where each subsequent bracket has \(2\) terms more than the previous bracket, find

1. the total number of terms in the first \(n\) brackets.[2]
2. the middle term of the \({{11}^{\mathrm{th}}}\) bracket in terms of \(a\).[2]
3. the range of values of \(a\) for the sum to infinity of the series to exist.
   Hence, find the least value of \(n\) for the sum of all the terms in the first \(n\) brackets to be within \(0.1\%\) of the sum to infinity of the series when \(a=\frac{39}{20}\).[4]