If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval :
(1) (– ∞

Q: Wait, why do we need the condition $|r| < 1$ at all? Can't we just plug numbers into the formula?

That is a fantastic question! The formula $S_{\infty} = \frac{a}{1-r}$ is derived from the limit of the partial sum $S_n = \frac{a(1-r^n)}{1-r}$ as $n$ approaches infinity. If $|r| \geq 1$, the term $r^n$ does not vanish; it either grows indefinitely or oscillates, meaning the sum does not converge to a single finite value. Without $|r| < 1$, the formula is mathematically meaningless!

Q: Why did the inequality signs flip when I multiplied by $-5$?

This is a classic trap! In algebra, whenever you multiply or divide an inequality by a negative number, the direction of the inequality must reverse. Think of it like a mirror: if $2 -5$. Always keep this in mind to avoid sign errors.

Q: Can $b$ be zero? The interval is $(0, 10)$, so zero is excluded.

Excellent observation. If $b = 0$, the first term is zero, and every subsequent term in the G.P. would also be zero. While the sum would technically be zero, a G.P. is generally defined with a non-zero first term to avoid triviality. In the context of this problem, $b=0$ would make the sum $0$, not $5$, so it must be excluded.

Q: Is it possible for $b$ to be negative?

If $b$ were negative, the sum $S_{\infty} = 5$ would require the common ratio $r$ to be greater than $1$ to make the denominator $(1-r)$ negative, which would violate the convergence condition $|r| < 1$. Thus, for a positive sum with a positive result, $b$ must remain within the positive bounds we derived.

Q: What happens if $b=10$? Does the sum become infinite?

If $b=10$, then $r = 1 - \frac{10}{5} = 1 - 2 = -1$. When $r = -1$, the series becomes $10, -10, 10, -10, \dots$. This is an oscillating series that does not converge to a single sum. Therefore, $b=10$ is the boundary where the series breaks down, which is why we use an open interval $(0, 10)$.

Question

If $b$ is the first term of an infinite G.P. whose sum is five, then $b$ lies in the interval :

Options:

(A) $(-\infty, -10]$

(B) $(-10, 0)$

(C) $(0, 10)$

(D) $[10, \infty)$

Sigma Percentile · Accepted Answer

### The Infinite Dance of Numbers Imagine you are standing at the edge of a vast, mathematical ocean. You have been tasked with summing an infinite number of terms—a Geometric Progression (G.P.)—and you are told that despite the infinite nature of this sequence, the total sum is exactly $5$. How is this possible? How can an infinite collection of numbers settle into a finite, elegant value? This is the beauty of the infinite G.P., and today, we are going to unlock the secret of its first term, $b$. ### The Convergence Condition: The Gatekeeper Before we dive into the algebra, we must respect the gatekeeper of infinite series: the convergence condition. For any infinite G.P. with first term $a$ and common ratio $r$, the sum $S_{\infty}$ is defined as: $$S_{\infty} = \frac{a}{1-r}$$ This formula is not a universal truth; it is a conditional one. It only holds power if the common ratio $r$ satisfies the condition $|r| < 1$. If $|r|$ were $1$ or greater, the terms would not shrink; they would either stay the same or grow, causing the sum to explode toward infinity or oscillate wildly. We must ensure our common ratio stays within the "safe zone" of $(-1, 1)$. ### Translating the Problem into Algebra We are given the first term $a = b$ and the sum $S_{\infty} = 5$. Let us substitute these into our formula: $$5 = \frac{b}{1-r}$$ Our goal is to understand the behavior of $b$. To do this, we need to isolate $r$ so we can apply our convergence condition. Rearranging the equation, we find: $$1 - r = \frac{b}{5}$$ $$r = 1 - \frac{b}{5}$$ This expression for $r$ is the key to the entire problem. It tells us exactly how the common ratio depends on our starting term $b$. ### The Inequality Journey Now, we invoke the gatekeeper. Since we know that $-1 < r < 1$, we substitute our expression for $r$ into this inequality: $$-1 < 1 - \frac{b}{5} < 1$$ First, let us subtract $1$ from all parts of the inequality to isolate the term containing $b$: $$-1 - 1 < -\frac{b}{5} < 1 - 1$$ $$-2 < -\frac{b}{5} < 0$$ Now, we face the most critical moment of the calculation. We need to isolate $b$ by multiplying the entire inequality by $-5$. Remember the golden rule of inequalities: whenever you multiply by a negative number, the inequality signs must flip their direction. It is a moment of transformation where the less-than signs become greater-than signs: $$(-2)(-5) > b > (0)(-5)$$ $$10 > b > 0$$ ### The Final Revelation We have arrived at our destination. The value of $b$ is constrained by the physical reality of the infinite series. It must be greater than $0$ and less than $10$. In the language of intervals, we write this as **$b \in (0, 10)$**. Look at what we have achieved! We started with a vague notion of an infinite sum and, through the rigor of the convergence condition and the careful manipulation of inequalities, we have defined the exact "territory" where $b$ is allowed to exist. This is the essence of JEE mathematics—not just calculating, but understanding the boundaries of possibility.

Animated Solution for Mathematics - Sequence and Series: If $b$ is the first term of an infinite G.P. whose sum is five, then $b$ lies in the interval :

Select Answer:

Visualized Solution

Identify the Infinite G.P. Parameters

Recall the Infinite Sum Formula

Substitute Given Values

Isolate the Common Ratio $r$

Apply Convergence Condition

Substitute $r$ in the Inequality

Simplify the Inequality - Step 1

Simplify the Inequality - Step 2

Final Interval for $b$

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

The Infinite Dance of Numbers

The Convergence Condition

Translating the Problem into Algebra

The Inequality Journey

The Final Revelation

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If b is the first term of an infinite G.P. whose sum is five, then b lies in the interval :