1. Key Concepts and Formulas

• Arithmetic Progression (A.P.):
  • The $n^{\text{th}}$ term is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
  • The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a_1 + (n-1)d]$.
• Geometric Progression (G.P.):
  • If three terms $A, B, C$ are in G.P., then the middle term is the geometric mean of the other two, i.e., $B^2 = AC$.

2. Step-by-Step Solution

Step 1: Define the terms of the A.P. and G.P. Let the first term of the A.P. be $a_1$ and its common difference be $d$. We are given that $a_1 = 1$. The problem states that the $2^{\text{nd}}$, $8^{\text{th}}$, and $44^{\text{th}}$ terms of this A.P. are the $1^{\text{st}}$, $2^{\text{nd}}$, and $3^{\text{rd}}$ terms of a G.P., respectively. Let these G.P. terms be $G_1, G_2, G_3$.

Using the A.P. formula, we express these terms:

• $G_1 = a_2 = a_1 + (2-1)d = 1 + d$
• $G_2 = a_8 = a_1 + (8-1)d = 1 + 7d$
• $G_3 = a_{44} = a_1 + (44-1)d = 1 + 43d$

Step 2: Apply the G.P. condition to find the common difference ($d$) of the A.P. Since $G_1, G_2, G_3$ are in G.P., the square of the middle term is equal to the product of the other two: $G_2^2 = G_1 \cdot G_3$. Substituting the expressions from Step 1: $(1 + 7d)^2 = (1 + d)(1 + 43d)$

Now, we expand both sides of the equation:

• Left side: $(1 + 7d)^2 = 1^2 + 2(1)(7d) + (7d)^2 = 1 + 14d + 49d^2$
• Right side: $(1 + d)(1 + 43d) = 1(1) + 1(43d) + d(1) + d(43d) = 1 + 43d + d + 43d^2 = 1 + 44d + 43d^2$

Equating the expanded forms: $1 + 14d + 49d^2 = 1 + 44d + 43d^2$

Step 3: Solve the quadratic equation for $d$. Rearrange the terms to form a standard quadratic equation: $49d^2 - 43d^2 + 14d - 44d + 1 - 1 = 0$ $6d^2 - 30d = 0$

Factor out the common term $6d$: $6d(d - 5) = 0$

This gives two possible values for $d$:

• $6d = 0 \implies d = 0$
• $d - 5 = 0 \implies d = 5$

The problem states that the A.P. is "non-constant".

• If $d=0$, all terms of the A.P. would be equal to the first term ($1$), making it a constant A.P. This contradicts the given condition.
• Therefore, we must discard $d=0$.

The common difference of the A.P. is $d=5$.

Step 4: Calculate the sum of the first 20 terms of the A.P. We need to find $S_{20}$ for the A.P. with $a_1 = 1$ and $d = 5$. Using the sum formula $S_n = \frac{n}{2}[2a_1 + (n-1)d]$: $S_{20} = \frac{20}{2}[2(1) + (20-1)(5)]$ $S_{20} = 10[2 + (19)(5)]$ $S_{20} = 10[2 + 95]$ $S_{20} = 10[97]$ $S_{20} = 970$

3. Common Mistakes and Tips

• Non-constant A.P. condition: Always check if the obtained common difference results in a constant A.P. If it does, discard that solution as it violates the problem statement.
• Algebraic accuracy: Double-check expansions and simplifications of algebraic expressions, especially when dealing with squares of binomials and products of binomials. Small errors can lead to incorrect values for $d$.
• Verification: After finding the value of $d$, it's a good practice to verify if the three terms indeed form a G.P. For $d=5$, the terms are $1+5=6$, $1+7(5)=36$, and $1+43(5)=216$. Since $36^2 = 1296$ and $6 \times 216 = 1296$, these terms are in G.P.

4. Summary

The problem requires us to connect terms of an Arithmetic Progression with terms of a Geometric Progression. We first expressed the relevant terms of the A.P. in terms of its first term ($a_1=1$) and common difference ($d$). These expressions were then used in the property of a G.P. ($B^2=AC$) to form an equation in $d$. Solving this equation, and considering the "non-constant" constraint, yielded the unique common difference of the A.P. Finally, the sum of the first 20 terms of this A.P. was calculated using the standard sum formula.

The final answer is $\boxed{970}$.