The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its ter

Q: Wait, why do we reject the root $r = 3/2$?

Great question! The sum of an infinite geometric series $S = \frac{a}{1-r}$ is only valid when $|r| < 1$. If $r = 3/2$, the terms would grow larger instead of smaller, and the sum would diverge to infinity. Always check your domain constraints—they are the silent guardians of your solution!

Q: Can we use the sum of cubes formula directly without expanding?

You could, but it is much harder. Expanding $1-r^3$ into $(1-r)(1+r+r^2)$ is a classic JEE technique to reveal hidden factors. It turns a cubic equation into a manageable quadratic one, saving you from unnecessary complexity.

Q: Is there a shortcut using the ratio of sums?

While you can manipulate the ratios, the substitution method is the most robust. It ensures you do not lose any information and keeps the algebra clean. Stick to the substitution method for reliability and to avoid algebraic traps.

Question

The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is :

Options:

(A) $\frac{4}{9}$

(B) $\frac{2}{9}$

(C) $\frac{2}{3}$

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a cliff, looking at a series of steps that get smaller and smaller, approaching a limit. This is the essence of an infinite geometric progression.

We start with a first term $a$ and a common ratio $r$. For the sum to exist, we must have $|r| < 1$. If $r$ were larger, our sum would explode to infinity, and the problem would lose its meaning.

We are given that the sum of this series is $3$. Using the formula $S = \frac{a}{1-r}$, we can write:

$$a = 3(1-r)$$

This is our first anchor.

### The Twist: Cubing the Terms

Now, the problem introduces a fascinating twist. What if we cube every single term? Our new series becomes $a^3, (ar)^3, (ar^2)^3, \dots$.

This is still a geometric progression, but with a new first term $A = a^3$ and a new common ratio $R = r^3$. The sum of this new series is given as $\frac{27}{19}$.

Applying the sum formula again, we get:

$$\frac{a^3}{1-r^3} = \frac{27}{19}$$

This is where the magic happens.

### The Algebraic Symphony

We have two equations, and it is time to bring them together. By substituting $a = 3(1-r)$ into our second equation, we get:

$$\frac{[3(1-r)]^3}{1-r^3} = \frac{27}{19}$$

Expanding the numerator, we have:

$$\frac{27(1-r)^3}{1-r^3} = \frac{27}{19}$$

We can immediately cancel the $27$ from both sides. Now, we face the expression:

$$\frac{(1-r)^3}{1-r^3} = \frac{1}{19}$$

Remember the identity $1-r^3 = (1-r)(1+r+r^2)$. Substituting this into the denominator, we get:

$$\frac{(1-r)^3}{(1-r)(1+r+r^2)} = \frac{1}{19}$$

Since $r 
eq 1$, we can safely cancel the $(1-r)$ factor, leaving us with:

$$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$$

### The Quadratic Finale

Now, we cross-multiply to clear the fractions:

$$19(1-r)^2 = 1+r+r^2$$

Expanding the left side, we get $19(1 - 2r + r^2) = 1 + r + r^2$, which simplifies to $19 - 38r + 19r^2 = 1 + r + r^2$. Bringing everything to one side, we arrive at the quadratic equation:

$$18r^2 - 39r + 18 = 0$$

Dividing by $3$, we get $6r^2 - 13r + 6 = 0$. Factorizing this, we find:

$$(3r - 2)(2r - 3) = 0$$

This gives us two potential roots: $r = \frac{2}{3}$ and $r = \frac{3}{2}$. Recalling our initial constraint that $|r| < 1$, we must reject $\frac{3}{2}$.

Thus, the only valid common ratio is **$r = \frac{2}{3}$**. You have successfully navigated the infinite, simplified the complex, and arrived at the truth.

Animated Solution for Mathematics - Sequence and Series: The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is $\frac{27}{19}$ . Then the common ratio of this series is :

Select Answer:

Visualized Solution

Defining the Infinite G.P.

Sum of the Series $S = 3$

Sum of Cubes $S^{'} = \frac{27}{19}$

Substitution Strategy

Simplifying the Equation

Canceling Common Factors

Cross Multiplication

Forming the Quadratic Equation

Simplifying and Factorizing

Finding the Common Ratio

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

Analyzing the Setup

The Twist

The Algebraic Symphony

The Quadratic Finale

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