Key Concepts and Formulas

• A Geometric Progression (G.P.) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The terms are $a, ar, ar^2, ar^3, \dots$.
• The $n$-th term of a G.P. is $T_n = ar^{n-1}$.
• The sum of the first $n$ terms of a G.P. is $S_n = a \frac{r^n - 1}{r - 1}$ (for $r \neq 1$).
• The reciprocals of the terms of a G.P. also form a G.P. with the first term $1/a$ and common ratio $1/r$.

Step-by-Step Solution

1. Representing the G.P. and Setting Up Equations Let the first term of the G.P. be $a$ and the common ratio be $r$. The first four terms are $a, ar, ar^2, ar^3$.

We are given the following information:

• The sum of the first four terms is $\frac{65}{12}$: $a + ar + ar^2 + ar^3 = \frac{65}{12}$ Factoring out $a$: $a(1 + r + r^2 + r^3) = \frac{65}{12} \quad \text{.........(1)}$

• The sum of their respective reciprocals is $\frac{65}{18}$: The reciprocals are $\frac{1}{a}, \frac{1}{ar}, \frac{1}{ar^2}, \frac{1}{ar^3}$. $\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} = \frac{65}{18}$ To sum these, we find a common denominator ($ar^3$): $\frac{r^3 + r^2 + r + 1}{ar^3} = \frac{65}{18}$ Rearranging the numerator: $\frac{1 + r + r^2 + r^3}{ar^3} = \frac{65}{18} \quad \text{.........(2)}$

• The product of the first three terms is 1: $a \cdot ar \cdot ar^2 = 1$ $a^3 r^3 = 1 \quad \text{.........(3)}$

2. Solving the System of Equations

From Equation (3), $a^3 r^3 = 1$, we can take the cube root of both sides to get $ar = 1$. This is a crucial relationship between $a$ and $r$.

Now, let's use Equations (1) and (2). Notice that both equations contain the term $(1 + r + r^2 + r^3)$. Dividing Equation (1) by Equation (2) will eliminate this common factor and simplify the problem.

$\frac{a(1 + r + r^2 + r^3)}{\frac{1 + r + r^2 + r^3}{ar^3}} = \frac{65/12}{65/18}$

Simplifying the left side by multiplying by the reciprocal of the denominator: $a \cdot (ar^3) = \frac{65}{12} \cdot \frac{18}{65}$

$a^2 r^3 = \frac{18}{12}$

$a^2 r^3 = \frac{3}{2} \quad \text{.........(4)}$

We have $ar = 1$. We can rewrite Equation (4) as $a \cdot (ar^3) = \frac{3}{2}$. Alternatively, we can write $a^2 r^3$ as $a \cdot (ar^2) \cdot r$. A more direct approach is to rewrite $a^2 r^3$ as $a \cdot (ar^3)$. However, we can also use $ar=1$ to substitute.

Let's use the relationship $ar=1$. From this, we have $a = 1/r$. Substitute this into Equation (4): $\left(\frac{1}{r}\right)^2 r^3 = \frac{3}{2}$ $\frac{1}{r^2} r^3 = \frac{3}{2}$ $r = \frac{3}{2}$

Now, using $ar=1$, we can find $a$: $a \left(\frac{3}{2}\right) = 1$ $a = \frac{2}{3}$

So, the first term is $a = \frac{2}{3}$ and the common ratio is $r = \frac{3}{2}$.

3. Finding the Third Term ($\alpha$) and $2\alpha$

The problem states that the third term of the G.P. is $\alpha$. The third term is given by $T_3 = ar^2$. Substituting the values of $a$ and $r$: $\alpha = \left(\frac{2}{3}\right) \left(\frac{3}{2}\right)^2$ $\alpha = \left(\frac{2}{3}\right) \left(\frac{9}{4}\right)$ $\alpha = \frac{2 \cdot 9}{3 \cdot 4} = \frac{18}{12}$ $\alpha = \frac{3}{2}$

The question asks for the value of $2\alpha$. $2\alpha = 2 \cdot \left(\frac{3}{2}\right)$ $2\alpha = 3$

Common Mistakes & Tips

• Algebraic Errors: Carefully handle fractions and exponents. When dividing equations, ensure you are multiplying by the reciprocal correctly.
• Product of Terms: For a G.P. with an odd number of terms, using terms like $a/r, a, ar$ can simplify the product calculation. However, for this problem, the standard representation $a, ar, ar^2, ar^3$ is perfectly adequate.
• Reciprocal Sum: Remember that the reciprocals of a G.P. also form a G.P. The structure of the sum of reciprocals often mirrors the original sum, which can be exploited through division.

Summary

This problem involves a system of equations derived from the properties of a geometric progression. We used the sum of the first four terms, the sum of their reciprocals, and the product of the first three terms to establish relationships between the first term ($a$) and the common ratio ($r$). By strategically dividing the equations for the sums, we simplified the problem to find $a$ and $r$. The relationship $ar=1$ derived from the product of the first three terms was key. Finally, we calculated the third term ($\alpha$) and then the required value of $2\alpha$.

The final answer is $\boxed{3}$.