Let a, ar, ar², ... be an infinite G.P. If Σ(n=0 to ∞) ar^n = 57 and Σ(n=0 to ∞) a³r^(3n) = 9747, th

Q: Wait, why do we reject $r = 3/2$? It satisfies the quadratic equation!

That is a brilliant observation! In the world of pure algebra, $r = 3/2$ is a perfectly valid root. However, we are working within the physical reality of an infinite geometric progression. For the sum to converge to a finite value, the common ratio must satisfy $|r| < 1$. Since $3/2 = 1.5$, which is greater than $1$, this value would cause the series to diverge to infinity. We reject it not because the math is wrong, but because it violates the physical constraints of the problem.

Q: Can I just use the formula for the sum of cubes directly?

Yes, you absolutely can! If you have a G.P. with first term $a$ and ratio $r$, the cubed series is a new G.P. with first term $a^3$ and ratio $r^3$. The sum formula $\frac{A}{1-R}$ still applies perfectly. The key is to correctly identify the new parameters $A = a^3$ and $R = r^3$.

Q: Is there a faster way than solving the quadratic?

While you might be tempted to look for shortcuts, the quadratic equation is the most robust path here. Once you simplify the constants, the quadratic $6r^2 - 13r + 6 = 0$ is quite elegant and factors easily. Trust the process—sometimes the most reliable path is the standard one!

Question

Let $a, ar, ar^2, \dots$ be an infinite G.P. If $\sum_{n=0}^\infty ar^n = 57$ and $\sum_{n=0}^\infty a^3 r^{3n} = 9747$, then $a + 18r$ is equal to

Options:

(A) 46

(B) 38

(C) 31

(D) 27

Sigma Percentile · Accepted Answer

### The Infinite Dance of Numbers

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are unraveling the secrets of an infinite geometric progression.

Imagine standing on the edge of an infinite sum. It feels daunting, but in the world of mathematics, infinity is not a wall; it is a gateway.

We are given two series: the original infinite G.P. and a second series formed by cubing every term of the first. Our goal is to find the value of $a + 18r$.

### The Guardian of Convergence

Before we touch a single variable, we must acknowledge the guardian of our problem: the convergence constraint. For any infinite geometric progression to have a finite sum, the common ratio $r$ must satisfy the condition $|r| < 1$.

If $r$ were to exceed this boundary, the series would explode toward infinity, and our sum of $57$ would be meaningless. Keep this constraint in your pocket; it will be the final judge of our solutions.

### The Transformation

We start with the sum of the first series:

$$ \frac{a}{1-r} = 57 $$

This gives us a beautiful, simple relationship: $a = 57(1-r)$.

Now, consider the second series. If our original terms are $a, ar, ar^2, \dots$, then the cubed terms are $a^3, a^3r^3, a^3r^6, \dots$. This is still a geometric progression with first term $A = a^3$ and common ratio $R = r^3$.

The sum of this new series is given as $9747$. Thus, we have our second equation:

$$ \frac{a^3}{1-r^3} = 9747 $$

### The Algebraic Symphony

Now, we perform the substitution. We know $a = 57(1-r)$, so $a^3 = 57^3(1-r)^3$. Substituting this into our second equation yields:

$$ \frac{57^3(1-r)^3}{1-r^3} = 9747 $$

Here is where the magic happens. The expression $1-r^3$ is a classic difference of cubes, which factors into $(1-r)(1+r+r^2)$.

When we substitute this back, we get:

$$ \frac{57^3(1-r)^3}{(1-r)(1+r+r^2)} = 9747 $$

We can safely cancel the $(1-r)$ term from the numerator and denominator because $r 
eq 1$. This leaves us with:

$$ \frac{57^3(1-r)^2}{1+r+r^2} = 9747 $$

Dividing $57^3$ by $9747$ simplifies perfectly to $19$. We are left with the elegant equation:

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

### The Final Resolution

Expanding this, we get $19(1 - 2r + r^2) = 1 + r + r^2$. Distributing the $19$ and rearranging terms leads us to the quadratic equation:

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

Dividing by $3$, we find $6r^2 - 13r + 6 = 0$. Factoring this, we get $(3r-2)(2r-3) = 0$.

This gives us two candidates for $r$: $2/3$ and $3/2$. Remember our guardian? Since $3/2 > 1$, we must reject it. Thus, $r = 2/3$ is our only valid ratio.

Substituting this back into our first equation:

$$ a = 57\left(1 - \frac{2}{3}ight) = 19 $$

Finally, we calculate the requested value:

$$ a + 18r = 19 + 18\left(\frac{2}{3}ight) = 19 + 12 = \mathbf{31} $$

We have arrived at our destination. The beauty of this problem lies not just in the answer, but in the way the complex expressions collapsed into simplicity.

Animated Solution for Mathematics - Sequence and Series: Let $a, a r, a r^{2}, \dots$ be an infinite G.P. If $\sum_{n = 0}^{\infty} a r^{n} = 57$ and $\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747$ , then $a + 18 r$ is equal to

Select Answer:

Visualized Solution

Given Conditions of the Infinite G.P.

Sum of the First Series

Sum of the Cubed Series

Substitution of $a^{3}$

Factorizing the Denominator

Simplifying the Constants

Expanding and Cross-Multiplying

Forming the Quadratic Equation

Solving for $r$

Selecting the Valid Common Ratio

Calculating the First Term $a$

Final Calculation

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

The Infinite Dance of Numbers

The Guardian of Convergence

The Transformation

The Algebraic Symphony

The Final Resolution

Similar Questions

If the sum of an infinite $GP$ $a, a r, a r^{2}, a r^{3}, \dots$ is $15$ and the sum of the squares of its each term is $150$ , then the sum of $a r^{2}, a r^{4}, a r^{6}, \dots$ is :

Let $A_{1}, A_{2}, A_{3}, \dots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$ , then, the value of $A_{6} + A_{8} + A_{10}$ is equal to

For the two positive numbers $a, b$ , if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then, $16 a + 12 b$ is equal to

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ......... .

Sum of infinite number of terms of GP is $20$ and sum of their square is $100$ . The common ratio of GP is

If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

Let $a_{1}, a_{2}, a_{3}, ...$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ then $a_{6}$ is equal to:

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

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; } Let a,ar,ar2,… be an infinite G.P. If ∑n=0∞​arn=57 and ∑n=0∞​a3r3n=9747, then a+18r is equal to