Key Concepts and Formulas

• Geometric Progression (G.P.): 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 typically represented as $a, ar, ar^2, \dots$.
• Arithmetic Progression (A.P.): A sequence where the difference between consecutive terms is constant (common difference, $d$). For three terms $A, B, C$ in A.P., the property $2B = A + C$ holds.
• Conditions for G.P.: For a G.P. with positive terms ($a>0$), an increasing G.P. implies that the common ratio $r$ must be greater than 1 ($r>1$). A decreasing G.P. implies $0 < r < 1$.

Step-by-Step Solution

Step 1: Represent the terms of the G.P. Let the three positive numbers forming an increasing G.P. be $a$, $ar$, and $ar^2$. Since the numbers are positive, $a > 0$. Since the G.P. is increasing, the common ratio $r$ must be greater than 1 ($r > 1$).

Step 2: Form the new sequence after doubling the middle term. According to the problem statement, if the middle term ($ar$) of the G.P. is doubled, the new numbers are in A.P. The new sequence is: $a$, $2ar$, $ar^2$.

Step 3: Apply the A.P. property to the new sequence. For the sequence $a, 2ar, ar^2$ to be in A.P., the middle term must be the arithmetic mean of the other two. This means: $2 \times (\text{middle term}) = (\text{first term}) + (\text{last term})$ $2(2ar) = a + ar^2$

Step 4: Simplify the equation and solve for $r$. $4ar = a + ar^2$ Since $a$ is a positive number ($a \neq 0$), we can divide the entire equation by $a$: $\frac{4ar}{a} = \frac{a}{a} + \frac{ar^2}{a}$ $4r = 1 + r^2$ Rearrange the terms to form a standard quadratic equation: $r^2 - 4r + 1 = 0$

Step 5: Solve the quadratic equation using the quadratic formula. The quadratic formula for an equation of the form $Ax^2 + Bx + C = 0$ is $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$. Here, $r$ is the variable, $A=1$, $B=-4$, and $C=1$. $r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)}$ $r = \frac{4 \pm \sqrt{16 - 4}}{2}$ $r = \frac{4 \pm \sqrt{12}}{2}$ Simplify $\sqrt{12}$: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$. $r = \frac{4 \pm 2\sqrt{3}}{2}$ Divide both terms in the numerator by 2: $r = 2 \pm \sqrt{3}$ This gives two possible values for $r$: $r_1 = 2 + \sqrt{3}$ and $r_2 = 2 - \sqrt{3}$.

Step 6: Select the correct value of $r$ based on the problem conditions. The problem states that the original G.P. is increasing and consists of positive numbers. For positive numbers, an increasing G.P. requires the common ratio $r > 1$. Let's evaluate the two possible values of $r$:

• $r_1 = 2 + \sqrt{3}$: Since $\sqrt{3} \approx 1.732$, $r_1 \approx 2 + 1.732 = 3.732$. This value is greater than 1, so it satisfies the condition for an increasing G.P.
• $r_2 = 2 - \sqrt{3}$: $r_2 \approx 2 - 1.732 = 0.268$. This value is less than 1, so it would correspond to a decreasing G.P. if the terms were positive.

Therefore, to satisfy the condition of an increasing G.P., the common ratio must be $r = 2 + \sqrt{3}$.

Common Mistakes & Tips

• Interpreting "Increasing G.P.": For positive terms, remember that an increasing G.P. implies $r>1$. This is a critical condition for filtering the correct solution.
• Dividing by 'a': Ensure that $a$ is not zero before dividing. In this problem, $a$ is stated to be a positive number, so $a \neq 0$, making division valid.
• Simplifying Radicals: Always simplify square roots (like $\sqrt{12} = 2\sqrt{3}$) to present the answer in its simplest form.

Summary

We represented the three positive numbers of the increasing G.P. as $a, ar, ar^2$. When the middle term was doubled, the new sequence $a, 2ar, ar^2$ formed an A.P. Using the A.P. property $2B = A+C$, we derived the quadratic equation $r^2 - 4r + 1 = 0$. Solving this equation gave two potential values for the common ratio: $2 + \sqrt{3}$ and $2 - \sqrt{3}$. Given that the original G.P. is increasing, the common ratio must be greater than 1. Thus, $r = 2 + \sqrt{3}$ is the correct common ratio.

The final answer is $\boxed{2 + \sqrt 3}$.