Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the

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

Great catch! In many physics problems, we reject negative roots because they don't make physical sense (like negative time or mass). However, in pure mathematics, we must be careful. In this specific problem, both roots of the quadratic equation are valid because both satisfy the condition $|r| < 1$. Always check your constraints before discarding a solution!

Q: Why can't we just guess the values? It seems faster.

Guessing is a dangerous game in JEE Advanced. While you might stumble upon one correct pair, you might miss the second valid solution. Algebra is the language of certainty. By setting up the system of equations, you guarantee that you find every possible solution, ensuring you don't fall into the trap of an incomplete answer.

Q: Why is the condition $|r| < 1$ so important?

Think of it as the 'gravity' of the series. If $|r| \ge 1$, the terms do not get smaller; they stay the same or grow larger. If you keep adding numbers that don't shrink, the sum will eventually explode to infinity. The condition $|r| < 1$ ensures that the terms shrink fast enough for the sum to settle down to a finite, beautiful number.

Question

Consider an infinite geometric series with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$, then

Options:

(A) $a = 4/7, r = 3/7$

(B) $a = 2, r = 3/8$

(C) $a = 3/2, r = 1/2$

(D) $a = 3, r = 1/4$

Sigma Percentile · Accepted Answer

### The Infinite Dance: Taming the Geometric Series

Welcome, future engineer. Today, we are not just solving a problem; we are taming infinity. There is something profoundly beautiful about an infinite geometric series.

Imagine standing on a line, taking a step of length $a$, then a step of length $ar$, then $ar^2$, and so on. If $r$ is small enough, you will never cross a certain point, no matter how many steps you take. That point is the sum of the series.

Let us dive into the mechanics of this convergence.

### Phase 1: The Infinite Sum Formula

We are given that the sum of an infinite geometric progression (G.P.) is $4$. The formula for the sum of an infinite G.P. is one of the most elegant results in algebra:

$$S_{\infty} = \frac{a}{1-r}$$

Here, $a$ is the first term and $r$ is the common ratio. This formula exists because of the condition $|r| < 1$. If the ratio were $1$ or greater, the terms would not shrink, and the sum would diverge.

By stating the sum is $4$, the problem implicitly tells us that our series is well-behaved and convergent. We can write our first equation:

$$\frac{a}{1-r} = 4$$

By multiplying both sides by $(1-r)$, we isolate the first term:

$$a = 4(1-r)$$

This is our first anchor. It tells us that $a$ and $r$ are locked in a dance; if you know one, you know the other.

### Phase 2: The Second Term Constraint

Now, we look at the second piece of information: the second term is $\frac{3}{4}$. Recall the general form of the $n^{th}$ term of a G.P.:

$$T_n = ar^{n-1}$$

For the second term ($n=2$), this simplifies beautifully to:

$$T_2 = ar$$

We are told this value is $\frac{3}{4}$. So, we have our second equation:

$$ar = \frac{3}{4}$$

We now have a system of two equations with two variables. This is the moment where many students panic, but you should feel excitement. We have all the information we need to force the variables to reveal themselves.

### Phase 3: The Algebraic Dance

Let us substitute our expression for $a$ from Phase 1 into our equation from Phase 2. We replace $a$ with $4(1-r)$:

$$4(1-r)r = \frac{3}{4}$$

Now, let us expand this. Distributing the $r$ gives us $4(r - r^2) = \frac{3}{4}$. To make our lives easier and eliminate the fraction, let us multiply the entire equation by $4$:

$$16(r - r^2) = 3$$

Expanding this further, we get $16r - 16r^2 = 3$. Rearranging everything to one side to form a standard quadratic equation, we arrive at:

$$16r^2 - 16r + 3 = 0$$

This is the heart of the problem. We have transformed a conceptual problem about infinity into a concrete quadratic equation.

### Phase 4: The Quadratic Resolution

Solving $16r^2 - 16r + 3 = 0$ requires us to split the middle term. We need two numbers that multiply to $(16 	imes 3) = 48$ and add to $-16$. Those numbers are $-12$ and $-4$.

So, we rewrite the equation:

$$16r^2 - 12r - 4r + 3 = 0$$

Grouping the terms, we factor out $4r$ from the first two and $-1$ from the last two:

$$4r(4r - 3) - 1(4r - 3) = 0$$

This gives us the factored form:

$$(4r - 1)(4r - 3) = 0$$

This yields two possible values for the common ratio:

$$r = \frac{1}{4} \quad 	ext{or} \quad r = \frac{3}{4}$$

Both values are less than $1$, so both are valid. Now, we find the corresponding $a$ values using $a = 4(1-r)$.

If $r = \frac{1}{4}$, then $a = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$. If $r = \frac{3}{4}$, then $a = 4(1 - \frac{3}{4}) = 4(\frac{1}{4}) = 1$.

### Conclusion

We have found two valid pairs: $(a=3, r=1/4)$ and $(a=1, r=3/4)$.

Looking at our options, we see that $(a=3, r=1/4)$ is listed. You have successfully navigated the constraints, set up the system, and solved the quadratic. This is the essence of JEE mathematics: taking a complex scenario and breaking it down into manageable, logical steps.

Keep this clarity of thought, and no problem will ever be too difficult for you.

Animated Solution for Mathematics - Sequence and Series: Consider an infinite geometric series with first term $a$ and common ratio $r$ . If its sum is $4$ and the second term is $3/4$ , then

Select Answer:

Visualized Solution

Visualizing the Infinite Series

The Infinite Sum Formula

Expressing $a$ in terms of $r$

Utilizing the Second Term

Substituting Equation One into Equation Two

Forming the Quadratic Equation

Solving the Quadratic Equation

Finding Corresponding Values of $a$

Matching with Options & Final Takeaway

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

The Infinite Dance

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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