1. Key Concepts and Formulas

   • Geometric Progression (GP): 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 ($r$). For four consecutive terms, they can be represented as $a, ar, ar^2, ar^3$.
   • Sum and Product of GP terms: For terms $a, ar, ar^2, ar^3$:
     • Sum ($S$) = $a(1+r+r^2+r^3)$
     • Product ($P$) = $a^4 r^6$
   • Algebraic Substitution for Symmetric Expressions: Expressions involving $x^n + x^{-n}$ can often be simplified by substituting $y = x^{1/2} + x^{-1/2}$. This leads to $y^2 = x + 2 + x^{-1}$, so $x + x^{-1} = y^2 - 2$. Similarly, $x^{3/2} + x^{-3/2}$ can be related to $y$.

2. Step-by-Step Solution

Step 1: Represent the terms and set up equations from the given information. Let the four positive consecutive terms of the G.P. be $a, ar, ar^2, ar^3$. Given that the terms are positive, we must have $a > 0$ and $r > 0$.

The product of the four terms is 1296: $(a)(ar)(ar^2)(ar^3) = 1296$ $a^4 r^{1+2+3} = 1296$ $a^4 r^6 = 1296$ Since $1296 = 6^4$, we have: $(ar^{3/2})^4 = 6^4$ Taking the positive fourth root (since $a, r > 0$): $ar^{3/2} = 6 \quad (*)$ From this, we can express $a$ in terms of $r$: $a = \frac{6}{r^{3/2}}$

The sum of the four terms is 126: $a + ar + ar^2 + ar^3 = 126$ Factor out $a$: $a(1 + r + r^2 + r^3) = 126$

Step 2: Substitute 'a' into the sum equation and simplify. Substitute $a = \frac{6}{r^{3/2}}$ into the sum equation: $\frac{6}{r^{3/2}}(1 + r + r^2 + r^3) = 126$ Divide both sides by 6: $\frac{1}{r^{3/2}}(1 + r + r^2 + r^3) = 21$ Distribute $\frac{1}{r^{3/2}}$: $\frac{1}{r^{3/2}} + \frac{r}{r^{3/2}} + \frac{r^2}{r^{3/2}} + \frac{r^3}{r^{3/2}} = 21$ Using exponent rules ($x^m/x^n = x^{m-n}$): $r^{-3/2} + r^{1 - 3/2} + r^{2 - 3/2} + r^{3 - 3/2} = 21$ $r^{-3/2} + r^{-1/2} + r^{1/2} + r^{3/2} = 21$ Group terms symmetrically: $(r^{3/2} + r^{-3/2}) + (r^{1/2} + r^{-1/2}) = 21$

Step 3: Use substitution to solve for 'r'. Let $y = r^{1/2} + r^{-1/2}$. We need to express $r^{3/2} + r^{-3/2}$ in terms of $y$. Consider the expansion of $y^3$: $y^3 = \left(r^{1/2} + r^{-1/2}\right)^3$ $y^3 = (r^{1/2})^3 + 3(r^{1/2})^2(r^{-1/2}) + 3(r^{1/2})(r^{-1/2})^2 + (r^{-1/2})^3$ $y^3 = r^{3/2} + 3r \cdot r^{-1/2} + 3r^{1/2} \cdot r^{-1} + r^{-3/2}$ $y^3 = r^{3/2} + 3r^{1/2} + 3r^{-1/2} + r^{-3/2}$ $y^3 = (r^{3/2} + r^{-3/2}) + 3(r^{1/2} + r^{-1/2})$ Substitute $y$ back into this equation: $y^3 = (r^{3/2} + r^{-3/2}) + 3y$ Therefore, $r^{3/2} + r^{-3/2} = y^3 - 3y$

Now substitute this and $y = r^{1/2} + r^{-1/2}$ into the equation from Step 2: $(y^3 - 3y) + y = 21$ $y^3 - 2y = 21$ $y^3 - 2y - 21 = 0$

Step 4: Solve the cubic equation for 'y'. We need to find the roots of the cubic equation $f(y) = y^3 - 2y - 21 = 0$. We can test integer factors of -21 (i.e., $\pm 1, \pm 3, \pm 7, \pm 21$). For $y=1$: $1^3 - 2(1) - 21 = 1 - 2 - 21 = -22 \neq 0$. For $y=3$: $3^3 - 2(3) - 21 = 27 - 6 - 21 = 0$. So, $y=3$ is a root. This means $(y-3)$ is a factor of the polynomial. We can perform polynomial division or synthetic division to find the other factors.

Dividing $y^3 - 2y - 21$ by $(y-3)$:

So, the cubic equation can be factored as $(y-3)(y^2 + 3y + 7) = 0$.

The roots are $y=3$ or $y^2 + 3y + 7 = 0$. For the quadratic equation $y^2 + 3y + 7 = 0$, the discriminant is $\Delta = b^2 - 4ac = 3^2 - 4(1)(7) = 9 - 28 = -19$. Since the discriminant is negative, the quadratic equation has no real roots for $y$.

Therefore, the only real solution for $y$ is $y=3$.

Step 5: Find the possible values of 'r' from the value of 'y'. We have $y = r^{1/2} + r^{-1/2} = 3$. Let $x = r^{1/2}$. Then $x + \frac{1}{x} = 3$. Multiply by $x$ (since $r>0$, $r^{1/2} \neq 0$): $x^2 + 1 = 3x$ $x^2 - 3x + 1 = 0$ Using the quadratic formula to solve for $x$: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)}$ $x = \frac{3 \pm \sqrt{9 - 4}}{2}$ $x = \frac{3 \pm \sqrt{5}}{2}$

Since $x = r^{1/2}$, we have $r^{1/2} = \frac{3 \pm \sqrt{5}}{2}$. Squaring both sides to find $r$: $r = \left(\frac{3 \pm \sqrt{5}}{2}\right)^2$ $r = \frac{3^2 \pm 2(3)\sqrt{5} + (\sqrt{5})^2}{2^2}$ $r = \frac{9 \pm 6\sqrt{5} + 5}{4}$ $r = \frac{14 \pm 6\sqrt{5}}{4}$ $r = \frac{7 \pm 3\sqrt{5}}{2}$

These are the two possible values for the common ratio $r$. Both values are positive, satisfying the condition that the terms are positive.

Step 6: Calculate the sum of all possible common ratios. The possible common ratios are $r_1 = \frac{7 + 3\sqrt{5}}{2}$ and $r_2 = \frac{7 - 3\sqrt{5}}{2}$. The sum of these common ratios is: $r_1 + r_2 = \frac{7 + 3\sqrt{5}}{2} + \frac{7 - 3\sqrt{5}}{2}$ $r_1 + r_2 = \frac{(7 + 3\sqrt{5}) + (7 - 3\sqrt{5})}{2}$ $r_1 + r_2 = \frac{7 + 3\sqrt{5} + 7 - 3\sqrt{5}}{2}$ $r_1 + r_2 = \frac{14}{2}$ $r_1 + r_2 = 7$

1. Common Mistakes & Tips

   • Assuming a different representation for GP terms: While $a, ar, ar^2, ar^3$ is standard, sometimes representing terms as $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ or similar symmetric forms can simplify product calculations. However, the given representation is also effective.
   • Errors in algebraic manipulation: Especially with fractional exponents and solving cubic equations. Carefully check each step.
   • Forgetting the condition of positive terms: Ensure that the derived common ratios lead to positive terms (which they do if $a>0$ and $r>0$).

2. Summary The problem involves finding the common ratios of a geometric progression given the sum and product of four consecutive positive terms. By setting up equations from the given information, expressing the first term in terms of the common ratio, and substituting into the sum equation, we obtained a solvable equation in terms of the common ratio. The use of a substitution $y = r^{1/2} + r^{-1/2}$ transformed a complex equation into a cubic polynomial. Solving this cubic equation yielded a real value for $y$, which was then used to find the possible values for the common ratio $r$. Finally, the sum of these possible common ratios was calculated.

The final answer is \boxed{7}. This corresponds to option (A).