If α and β are the roots of the equation 375x² - 25x - 2 = 0, then lim(n→∞) Σ(r=1 to n) α^r + lim(n→

Q: Wait, why do we reject the negative root here?

Actually, we don't reject any roots! Both $\alpha$ and $\beta$ are valid. The beauty of the symmetric expression $\frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta}$ is that it treats both roots equally. We don't need to know which is which, because the final result depends only on their sum and product.

Q: Why can't we just use energy conservation?

I love the enthusiasm! While this isn't a physics problem, in algebra, our 'conservation' is the conservation of symmetry. By using Vieta's formulas, we conserve the information about the roots without needing to calculate their messy individual values. It is the most elegant path to the truth.

Q: How do I know the series converges?

Great question! For an infinite geometric series $\sum r^n$ to converge, the common ratio must satisfy $|r| < 1$. Since the roots of $375x^2 - 25x - 2 = 0$ are small, they satisfy this condition, ensuring our series sums are finite and well-defined. Always check the convergence condition before summing to infinity!

Question

If $\alpha$ and $\beta$ are the roots of the equation $375x^2 - 25x - 2 = 0$, then $\lim_{n 	o \infty} \sum_{r=1}^n \alpha^r + \lim_{n 	o \infty} \sum_{r=1}^n \beta^r$ is equal to :

Options:

(A) $21/346$

(B) $29/358$

(C) $1/12$

(D) $7/116$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The equation $375x^2 - 25x - 2 = 0$ serves as our foundation. While one could solve for the roots $\alpha$ and $\beta$ directly, the JEE Advanced approach prioritizes efficiency and structural insight.

We rely on **Vieta's formulas** to extract the necessary information without calculating the roots explicitly.

For the quadratic equation $ax^2 + bx + c = 0$, the sum and product of the roots are given by:

$$\alpha + \beta = -\frac{b}{a} = -\frac{-25}{375} = \frac{1}{15}$$

$$\alpha\beta = \frac{c}{a} = -\frac{2}{375}$$

These two values, $\frac{1}{15}$ and $-\frac{2}{375}$, are the essential keys to solving the problem.

### Unmasking the Infinite

We are tasked with evaluating the expression $S = \lim_{n 	o \infty} \sum_{r=1}^n \alpha^r + \lim_{n 	o \infty} \sum_{r=1}^n \beta^r$. As $n$ approaches infinity, these represent the sums of two infinite geometric series.

The sum of an infinite geometric series is given by $S_{\infty} = \frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio. Applying this to our series:

$$S = \frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta}$$

### The Algebraic Dance

To simplify this expression, we combine the fractions over a common denominator:

$$S = \frac{\alpha(1-\beta) + \beta(1-\alpha)}{(1-\alpha)(1-\beta)}$$

Expanding the numerator yields $\alpha - \alpha\beta + \beta - \alpha\beta$, which simplifies to $(\alpha+\beta) - 2\alpha\beta$. The denominator expands to $1 - (\alpha+\beta) + \alpha\beta$.

Thus, the expression is entirely defined by the sum and product of the roots:

$$S = \frac{(\alpha+\beta) - 2\alpha\beta}{1 - (\alpha+\beta) + \alpha\beta}$$

### The Final Triumphant Calculation

Now, we substitute the values $\alpha+\beta = \frac{1}{15}$ and $\alpha\beta = -\frac{2}{375}$ into our derived formula.

The numerator becomes:

$$\frac{1}{15} - 2\left(-\frac{2}{375}ight) = \frac{25}{375} + \frac{4}{375} = \frac{29}{375}$$

The denominator becomes:

$$1 - \frac{1}{15} + \left(-\frac{2}{375}ight) = \frac{375 - 25 - 2}{375} = \frac{348}{375}$$

Finally, we calculate the ratio:

$$S = \frac{29/375}{348/375} = \frac{29}{348}$$

Since $29 	imes 12 = 348$, the final result is:

**$S = \frac{1}{12}$**

Animated Solution for Mathematics - Sequence and Series: If $α$ and $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ , then $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$ is equal to :

Select Answer:

Visualized Solution

Identify the Equation and its Roots

Calculate Sum of Roots

Calculate Product of Roots

Interpret the Infinite Summation

Apply Infinite G.P. Formula

Combine the Algebraic Fractions

Expand and Simplify the Expression

Substitute Known Values

Simplify the Numerator

Simplify the Denominator

Final Calculation and Simplification

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

Analyzing the Setup

Unmasking the Infinite

The Algebraic Dance

The Final Triumphant Calculation

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