Let {a_k} and {b_k}, k ∈ N, be two G.P.s with common ratio r₁ and r₂ respectively such that a₁ = b₁

Q: Wait, why do we need to form a quadratic equation? Can't we just substitute $r_2 = \frac{5}{4} - r_1$ into the second equation?

You absolutely can! That is a perfectly valid algebraic path. However, using the quadratic equation approach is a powerful tool in your JEE arsenal. It minimizes the risk of messy algebraic errors and is often faster. Think of it as choosing the most efficient path through a forest—both lead to the destination, but one is a paved road.

Q: How do I know for sure that the infinite sum $\sum_{k=1}^{\infty} c_k$ actually converges?

Excellent question! A geometric series $\sum ar^{k-1}$ converges if and only if the absolute value of the common ratio is less than 1. Here, our ratios are $\frac{1}{2}$ and $\frac{3}{4}$. Since both $|\frac{1}{2}| < 1$ and $|\frac{3}{4}| < 1$, both series converge beautifully. Always check this condition before applying the formula $S_{\infty} = \frac{a}{1-r}$!

Q: What if I accidentally swapped $r_1$ and $r_2$? Does it change the final answer?

In this specific problem, the final expression $\sum c_k - (12a_6 + 8b_4)$ is sensitive to which ratio belongs to which sequence. The problem explicitly states $r_1 < r_2$. If you swap them, you will calculate the wrong terms for $a_6$ and $b_4$. Always respect the constraints given in the problem statement—they are the guardrails of your solution.

Question

Let $\{a_k\}$ and $\{b_k\}, k \in \mathbb{N}$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1 = b_1 = 4$ and $r_1 < r_2$. Let $c_k = a_k + b_k, k \in \mathbb{N}$. If $c_2 = 5$ and $c_3 = \frac{13}{4}$ then $\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to

Sigma Percentile · Accepted Answer

### The Symphony of Sequences: A Journey into Geometric Progressions

Welcome, my dear student. Today, we are not just solving a problem; we are conducting a symphony. We have two geometric progressions, $\{a_k\}$ and $\{b_k\}$, dancing to the rhythm of their common ratios, $r_1$ and $r_2$.

When we define a new sequence $c_k = a_k + b_k$, we are essentially superimposing two waves. Our goal is to decode the hidden parameters of these sequences and find the value of a complex expression. Let us begin.

### Phase 1: Decoding the Clues

We start with the basics. We know the first terms $a_1 = 4$ and $b_1 = 4$. The beauty of a geometric progression lies in its simplicity: every term is just the first term multiplied by the ratio raised to a power.

We are given two snapshots of the sequence $c_k$: $c_2 = 5$ and $c_3 = \frac{13}{4}$. Let us translate these into the language of algebra. Since $c_k = a_k + b_k$, we have:

$$c_2 = a_2 + b_2 = 4r_1 + 4r_2 = 5$$

Dividing by 4, we get our first elegant relation: $r_1 + r_2 = \frac{5}{4}$. Next, for the third term:

$$c_3 = a_3 + b_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4}$$

Again, dividing by 4, we find: $r_1^2 + r_2^2 = \frac{13}{16}$.

### Phase 2: The Algebraic Bridge

Now, we stand at a crossroads. We have the sum of the ratios and the sum of their squares. This is a classic setup that screams for an algebraic identity.

We know that $(r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2$. This identity is the bridge that will take us from the sum to the product. Substituting our known values:

$$\left(\frac{5}{4}\right)^2 = \frac{13}{16} + 2r_1r_2$$

$$\frac{25}{16} = \frac{13}{16} + 2r_1r_2$$

Subtracting $\frac{13}{16}$ from both sides, we get $2r_1r_2 = \frac{12}{16}$, which simplifies to $2r_1r_2 = \frac{3}{4}$. Thus, the product of our ratios is $r_1r_2 = \frac{3}{8}$.

### Phase 3: The Quadratic Reveal

Here is where the magic happens. In the JEE Advanced, we often encounter the sum and product of two unknowns. Whenever you see this, think of a quadratic equation.

If $r_1$ and $r_2$ are roots of a quadratic equation $x^2 - Sx + P = 0$, where $S$ is the sum and $P$ is the product, then:

$$x^2 - \frac{5}{4}x + \frac{3}{8} = 0$$

To make this look cleaner, let us multiply by 8: $8x^2 - 10x + 3 = 0$. Factoring this quadratic is a joy. We look for numbers that multiply to 24 and add to -10.

Those are -6 and -4. So, $8x^2 - 6x - 4x + 3 = 0$, which factors into $(4x - 3)(2x - 1) = 0$. Our roots are $x = \frac{3}{4}$ and $x = \frac{1}{2}$. Given the constraint $r_1 < r_2$, we assign $r_1 = \frac{1}{2}$ and $r_2 = \frac{3}{4}$.

### Phase 4: The Infinite Sum and Final Calculation

We are nearing the finish line. We need to calculate $\sum_{k=1}^{\infty} c_k$. Since $c_k = a_k + b_k$, the sum of the infinite series is simply the sum of the two individual infinite geometric series.

The formula $S_{\infty} = \frac{a}{1-r}$ is our best friend here. For sequence $a_k$:

$$\sum a_k = \frac{4}{1 - \frac{1}{2}} = \frac{4}{\frac{1}{2}} = 8$$

For sequence $b_k$:

$$\sum b_k = \frac{4}{1 - \frac{3}{4}} = \frac{4}{\frac{1}{4}} = 16$$

So, $\sum c_k = 8 + 16 = 24$. Finally, we calculate the specific terms $12a_6$ and $8b_4$:

$$a_6 = 4\left(\frac{1}{2}\right)^5 = \frac{4}{32} = \frac{1}{8} \Rightarrow 12a_6 = \frac{12}{8} = \frac{3}{2}$$

$$b_4 = 4\left(\frac{3}{4}\right)^3 = 4\left(\frac{27}{64}\right) = \frac{27}{16} \Rightarrow 8b_4 = 8\left(\frac{27}{16}\right) = \frac{27}{2}$$

Putting it all together:

$$24 - \left(\frac{3}{2} + \frac{27}{2}\right) = 24 - \frac{30}{2} = 24 - 15 = 9$$

And there it is—the number **9**. A beautiful, clean result born from the interplay of sequences and algebra. You have mastered the logic. Keep this clarity, and you will conquer any problem the exam throws at you.

Enter Numerical Value:

Visualized Solution

Given Information

Setting up the First Equation

Setting up the Second Equation

Algebraic Identity for Product

Calculating $r_{1} r_{2}$

Forming the Quadratic Equation

Solving the Quadratic Equation

Assigning $r_{1}$ and $r_{2}$

Calculating Infinite Sums

Calculating Specific Terms

Final Evaluation

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

The Symphony of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

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