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Let's break down the problem and solve it step by step.

### Given Information:

- The 2nd, 8th, and 44th terms of an arithmetic progression (AP) form the first three terms of a geometric progression (GP).
- The first term of the AP is 1.
- The common difference of the AP is positive.

### Part (a)(i):

**We need to show that the common difference of the arithmetic progression is 5.**

Let the first term of the AP be denoted as \( a_1 = 1 \), and the common difference as \( d \).

The general term of an AP is given by:

\[
a_n = a_1 + (n-1) \cdot d
\]

So, we can find the 2nd, 8th, and 44th terms of the AP:

- \( a_2 = 1 + (2-1) \cdot d = 1 + d \)
- \( a_8 = 1 + (8-1) \cdot d = 1 + 7d \)
- \( a_{44} = 1 + (44-1) \cdot d = 1 + 43d \)

Since these three terms form a geometric progression, the condition for a sequence to be in a GP is:

\[
\frac{a_8}{a_2} = \frac{a_{44}}{a_8}
\]

Substitute the values for \( a_2 \), \( a_8 \), and \( a_{44} \):

\[
\frac{1 + 7d}{1 + d} = \frac{1 + 43d}{1 + 7d}
\]

Now, cross-multiply to solve this equation:

\[
(1 + 7d)^2 = (1 + d)(1 + 43d)
\]

Expand both sides:

\[
(1 + 7d)^2 = 1 + 14d + 49d^2
\]
\[
(1 + d)(1 + 43d) = 1 + 43d + d + 43d^2 = 1 + 44d + 43d^2
\]

Now set the two expressions equal to each other:

\[
1 + 14d + 49d^2 = 1 + 44d + 43d^2
\]

Cancel out the common terms:

\[
14d + 49d^2 = 44d + 43d^2
\]

Rearrange the equation:

\[
49d^2 - 43d^2 = 44d - 14d
\]
\[
6d^2 = 30d
\]

Divide through by \( d \) (since \( d \neq 0 \)):

\[
6d = 30
\]

Solve for \( d \):

\[
d = 5
\]

Thus, the common difference of the arithmetic progression is **5**.

---

### Part (a)(ii):

**We need to find the sum of the first 20 terms of the arithmetic progression.**

The formula for the sum \( S_n \) of the first \( n \) terms of an arithmetic progression is given by:

\[
S_n = \frac{n}{2} \left(2a_1 + (n-1) \cdot d \right)
\]

For the first 20 terms (\( n = 20 \)):

\[
S_{20} = \frac{20}{2} \left(2 \cdot 1 + (20-1) \cdot 5 \right)
\]
\[
S_{20} = 10 \left(2 + 19 \cdot 5 \right)
\]
\[
S_{20} = 10 \left(2 + 95 \right) = 10 \cdot 97 = 970
\]

Thus, the sum of the first 20 terms of the arithmetic progression is **970**.

---

Would you like more details or further clarification?

### Related Questions:
1. What is the formula for the \(n\)-th term of an arithmetic progression?
2. How can we verify if three terms form a geometric progression?
3. What is the relationship between the common difference and sum of an arithmetic progression?
4. Can the common difference in an arithmetic progression ever be negative?
5. How do we generalize this problem to other terms forming a geometric progression?

### Tip:
Always cross-check whether your solution satisfies the initial conditions of the problem (in this case, ensuring the common difference is positive and the terms form a valid progression).

The 2nd, 8th, and 44th terms of an arithmetic progression form the first three terms of a geometric progression. The first term is 1, and the common difference is positive. (i) Show that the common difference is 5. (ii) Find the sum of the first 20 terms of the arithmetic progression.

Learn how to find the common difference of an arithmetic progression given that certain terms form a geometric progression. This problem also covers how to calculate the sum of the first 20 terms. Suitable for high school students (Grades 10-12).

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