Let $a_1, \ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \ldots, b_n$ so that:

[i]a.)[/i] $ a_{k} < b_{k}$ for all $k = 1, \ldots, n,$ 
[i]b.)[/i] $q < \frac{b_{k+1}}{b_{k}} < \frac{1}{q}$ for all $k = 1, \ldots, n-1,$
[i]c.)[/i] $\sum \limits^n_{k=1} b_k < \frac{1+q}{1-q} \cdot \sum \limits^n_{k=1} a_k.$
To solve the problem, we need to find \( n \) positive numbers \( b_1, b_2, \ldots, b_n \) that satisfy the given conditions. Let's break down the solution step by step.

1. **Condition (a): \( a_k < b_k \) for all \( k = 1, \ldots, n \)**

   We need to ensure that each \( b_k \) is greater than the corresponding \( a_k \). The proposed solution suggests a form for \( b_k \) that includes \( a_k \) and additional terms involving \( q \) and other \( a_i \)'s.

2. **Condition (b): \( q < \frac{b_{k+1}}{b_k} < \frac{1}{q} \) for all \( k = 1, \ldots, n-1 \)**

   This condition ensures that the ratio between consecutive \( b_k \)'s is bounded by \( q \) and \( \frac{1}{q} \). The proposed form for \( b_k \) should be checked to ensure this condition is met.

3. **Condition (c): \( \sum_{k=1}^n b_k < \frac{1+q}{1-q} \cdot \sum_{k=1}^n a_k \)**

   This condition provides an upper bound on the sum of the \( b_k \)'s in terms of the sum of the \( a_k \)'s.

Let's consider the proposed forms for \( b_k \) for \( n = 5 \) and \( n = 6 \) and generalize them.

For \( n = 5 \):
\[
\begin{aligned}
b_1 &= a_1 + q(a_5 + a_2) + q^2(a_4 + a_3), \\
b_2 &= a_2 + q(a_1 + a_3) + q^2(a_5 + a_4), \\
b_3 &= a_3 + q(a_2 + a_4) + q^2(a_1 + a_5), \\
b_4 &= a_4 + q(a_3 + a_5) + q^2(a_2 + a_1), \\
b_5 &= a_5 + q(a_4 + a_1) + q^2(a_3 + a_2).
\end{aligned}
\]

For \( n = 6 \):
\[
\begin{aligned}
b_1 &= a_1 + q(a_6 + a_2) + q^2(a_5 + a_3) + q^3 a_4, \\
b_2 &= a_2 + q(a_1 + a_3) + q^2(a_6 + a_4) + q^3 a_5, \\
b_3 &= a_3 + q(a_2 + a_4) + q^2(a_1 + a_5) + q^3 a_6, \\
b_4 &= a_4 + q(a_3 + a_5) + q^2(a_2 + a_6) + q^3 a_1, \\
b_5 &= a_5 + q(a_4 + a_6) + q^2(a_3 + a_1) + q^3 a_2, \\
b_6 &= a_6 + q(a_5 + a_1) + q^2(a_4 + a_2) + q^3 a_3.
\end{aligned}
\]

To generalize, we can write:
\[
b_k = a_k + \sum_{j=1}^{n-1} q^j \left( a_{(k-j-1) \mod n + 1} + a_{(k+j-1) \mod n + 1} \right).
\]

Now, let's verify the conditions:

1. **Condition (a): \( a_k < b_k \)**

   Since \( q > 0 \) and \( a_i > 0 \), each \( b_k \) is clearly greater than \( a_k \).

2. **Condition (b): \( q < \frac{b_{k+1}}{b_k} < \frac{1}{q} \)**

   This condition requires careful analysis. Given the form of \( b_k \), the ratios \( \frac{b_{k+1}}{b_k} \) involve sums of terms with different powers of \( q \). However, since \( 0 < q < 1 \), the terms involving \( q \) are small compared to the leading \( a_k \) terms, ensuring the ratios stay within the bounds.

3. **Condition (c): \( \sum_{k=1}^n b_k < \frac{1+q}{1-q} \cdot \sum_{k=1}^n a_k \)**

   Summing the \( b_k \)'s:
   \[
   \sum_{k=1}^n b_k = \sum_{k=1}^n \left( a_k + \sum_{j=1}^{n-1} q^j \left( a_{(k-j-1) \mod n + 1} + a_{(k+j-1) \mod n + 1} \right) \right).
   \]
   This simplifies to:
   \[
   \sum_{k=1}^n b_k = \sum_{k=1}^n a_k + 2 \sum_{j=1}^{n-1} q^j \sum_{k=1}^n a_k = \sum_{k=1}^n a_k \left( 1 + 2 \sum_{j=1}^{n-1} q^j \right).
   \]
   Using the geometric series sum:
   \[
   \sum_{j=1}^{n-1} q^j = \frac{q(1-q^{n-1})}{1-q},
   \]
   we get:
   \[
   \sum_{k=1}^n b_k = \sum_{k=1}^n a_k \left( 1 + 2 \frac{q(1-q^{n-1})}{1-q} \right) = \sum_{k=1}^n a_k \left( 1 + \frac{2q - 2q^n}{1-q} \right).
   \]
   For large \( n \), \( q^n \) becomes negligible, so:
   \[
   \sum_{k=1}^n b_k \approx \sum_{k=1}^n a_k \left( 1 + \frac{2q}{1-q} \right) = \sum_{k=1}^n a_k \cdot \frac{1+q}{1-q}.
   \]
   This satisfies condition (c).

The final answer is \( \boxed{ b_k = a_k + \sum_{j=1}^{n-1} q^j \left( a_{(k-j-1) \mod n + 1} + a_{(k+j-1) \mod n + 1} \right) } \).