The sum of first four terms of a geometric progression (G.P.) is 65/12 and the sum of their respecti

Q: Why did we divide the equations? Couldn't we just solve for $a$ in terms of $r$ from the first equation and plug it into the second?

You absolutely could, but that is the path of maximum resistance! If you solve for $a$ from the first equation, you get $a = \frac{65}{12(1+r+r^2+r^3)}$. Substituting this into the second equation would lead to a high-degree polynomial in $r$ that is incredibly tedious to solve. In JEE Advanced, the 'trick' is often to look for symmetry. Dividing the equations allows the most complex part of the expression—the polynomial $(1+r+r^2+r^3)$—to vanish, simplifying the problem instantly. Always look for ways to cancel, not just substitute.

Q: Is it always true that the sum of reciprocals of a G.P. is related to the sum of the terms by the product of the first and last terms?

That is a brilliant observation! Yes, for a G.P. with $n$ terms, the sum of reciprocals is indeed the sum of the terms divided by the product of the first and last terms. Specifically, $\sum \frac{1}{a_i} = \frac{\sum a_i}{a_1 \cdot a_n}$. In our case, $a_1 = a$ and $a_4 = ar^3$, so the product is $a^2r^3$. This is a powerful property that can save you minutes of calculation in an exam setting.

Q: Wait, the question asks for $2\alpha$. Why is $\alpha$ just equal to $r$?

Let's re-examine the definition. The third term of our G.P. is $ar^2$. We found from the product condition that $a \cdot ar \cdot ar^2 = 1$, which simplifies to $(ar)^3 = 1$, meaning $ar = 1$. Now, look at the third term again: $ar^2 = (ar) \cdot r$. Since $ar = 1$, this becomes $1 \cdot r$, which is simply $r$. So yes, in this specific problem, the third term is numerically equal to the common ratio!

Question

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$. If the product of first three terms of the G.P. is 1, and the third term is $\alpha$, then $2\alpha$ is

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### The Elegance of Geometric Symmetry: A Journey Through Sequences

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden architecture of a Geometric Progression (G.P.).

When you first look at a problem involving sums and reciprocals, it is natural to feel a sense of dread. You might see fractions, powers, and variables, and your instinct might be to reach for the most complicated algebraic expansion possible.

But pause. Take a breath. In the world of JEE Advanced, the most complex-looking problems often hide the most beautiful, symmetrical solutions. Let us walk through this together.

### Phase 1: The Setup and the Hidden Symmetry

We begin by defining our sequence. A G.P. is defined by its first term, $a$, and its common ratio, $r$. Our four terms are $a$, $ar$, $ar^2$, and $ar^3$.

The problem gives us two distinct pieces of information. First, the sum of these terms is $\frac{65}{12}$. We write this as:

$$a(1 + r + r^2 + r^3) = \frac{65}{12} \quad \dots (1)$$

Now, consider the sum of their reciprocals. This is where many students stumble, trying to find a common denominator for the entire expression. But look at the structure: $\frac{1}{a} + \frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3}$.

If we factor out $\frac{1}{a}$ and find a common denominator of $r^3$, we get:

$$\frac{1}{a} \left( \frac{r^3 + r^2 + r + 1}{r^3} \right) = \frac{65}{18} \quad \dots (2)$$

Do you see it? The numerator inside the parenthesis is exactly the same as the factor in our first equation! This is the 'Aha!' moment. We have two equations that share a common, complex polynomial structure.

### Phase 2: The Power of Division

In algebra, we are often taught to substitute. But here, substitution is a trap. If we divide Equation (1) by Equation (2), the term $(1 + r + r^2 + r^3)$ will vanish entirely.

It is a moment of pure mathematical liberation. Watch the cancellation:

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

On the left side, the polynomial cancels out, and we are left with $a^2 r^3$. On the right side, the $65$ cancels out, and we are left with $\frac{18}{12}$, which simplifies to $\frac{3}{2}$.

Thus, we arrive at a remarkably simple relationship:

$$a^2 r^3 = \frac{3}{2} \quad \dots (3)$$

This is the power of looking for symmetry. We have reduced a daunting system of equations into a single, elegant relationship.

### Phase 3: Unlocking the Variables

Now, we turn to the third piece of information: the product of the first three terms is $1$. This is our key to unlocking the values of $a$ and $r$.

We write:

$$a \cdot (ar) \cdot (ar^2) = 1$$

This simplifies to $a^3 r^3 = 1$, or $(ar)^3 = 1$. Taking the cube root, we find that $ar = 1$.

This implies that $a = \frac{1}{r}$. This is a massive breakthrough. We now have a way to express $a$ entirely in terms of $r$. Let us substitute this into our previous result, $a^2 r^3 = \frac{3}{2}$:

$$\left( \frac{1}{r} \right)^2 \cdot r^3 = \frac{3}{2}$$

$$\frac{1}{r^2} \cdot r^3 = \frac{3}{2}$$

$$r = \frac{3}{2}$$

### Phase 4: The Final Victory

The problem asks for the value of $2\alpha$, where $\alpha$ is the third term of the G.P. We know the third term is $ar^2$.

Since we established that $ar = 1$, we can rewrite the third term as:

$$\alpha = (ar) \cdot r = 1 \cdot r = r$$

So, $\alpha = \frac{3}{2}$. The final step is to calculate $2\alpha$. It is crucial to stay focused here—do not stop at $\alpha$. We must multiply by $2$:

$$2\alpha = 2 \cdot \frac{3}{2} = 3$$

And there it is. The answer is **3**. We navigated the complexity, identified the symmetry, and used the constraints to peel back the layers of the problem.

Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the most elegant path to the truth. You have the tools; now, trust your intuition.

Animated Solution for Mathematics - Sequence and Series: The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

Enter Numerical Value:

Visualized Solution

Define the G.P. Terms

Sum of First Four Terms

Sum of Reciprocals

Simplify the Reciprocal Equation

Dividing the Equations

Product of First Three Terms

Solving for $r$

Finding the Third Term $α$

Final Calculation

The Sigma Insight: Geometric Progression (G.P.)

The Elegance of Geometric Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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