The sum of the infinite series 1 + 2/3 + 7/3² + 12/3³ + 17/3⁴ + 22/3⁵ + ... is equal to (1) 9/4 (2)

Q: Why do we ignore the first term when identifying the AP? The first term is $1$, but the second is $2/3$. The numerators are $1, 2, 7, 12...$ Wait, $1$ to $2$ is $+1$, but $2$ to $7$ is $+5$. The pattern starts from the second term.

That is a brilliant observation! You have correctly identified that this is a 'modified' AGP. The first term is an outlier that doesn't fit the pattern of the rest of the series. In such cases, we simply treat the first term as a separate entity and apply the AGP method to the infinite tail starting from the second term. Don't let the first term intimidate you; it's just a small hurdle before the beautiful pattern begins.

Q: Can we use the direct formula $S = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$? It seems much faster than the shift-and-subtract method.

You absolutely can use the formula, but be extremely careful! That formula is derived for a standard AGP starting from the first term. Because our series is modified, you would need to adjust it, which often leads to sign errors. The 'Shift and Subtract' method is much safer because it derives the result from first principles, ensuring you understand the 'why' behind every step. For JEE, conceptual clarity is your best defense against traps.

Q: Why does the infinite series converge? How do we know it doesn't just go to infinity?

Great question! The sum of an infinite GP converges if and only if the absolute value of the common ratio $|r| < 1$. In our series, the GP part has a common ratio of $r = \frac{1}{3}$. Since $|\frac{1}{3}| < 1$, the series is mathematically guaranteed to settle down to a finite value. If $r$ were greater than or equal to $1$, the series would indeed diverge to infinity, and we wouldn't be able to find a sum!

Question

The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \frac{17}{3^4} + \frac{22}{3^5} + \dots$ is equal to

Options:

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

(B) $\frac{15}{4}$

(C) $\frac{13}{4}$

(D) $\frac{11}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to peel back the layers of an infinite series that might look like a chaotic mess at first glance. We have the series:

$$S = 1 + \frac{2}{3} + \frac{7}{3^2} + \frac{12}{3^3} + \frac{17}{3^4} + \dots$$

When you see a series where each term is a product of an arithmetic term and a geometric term, you are looking at an Arithmetico-Geometric Progression, or AGP. Let us decode its DNA.

The numerators, starting from the second term, are $2, 7, 12, 17, \dots$. These form an Arithmetic Progression (AP) with a common difference $d = 5$.

The denominators are $3^1, 3^2, 3^3, \dots$, which form a Geometric Progression (GP) with a common ratio $r = \frac{1}{3}$. This is a 'modified' AGP because the first term, $1$, doesn't quite fit the pattern. But fear not! We have a universal tool for this: the Shift and Subtract method.

### The Art of the Shift

The beauty of the AGP method lies in its elegance. We take our sum $S$ and multiply it by the common ratio of the GP, which is $\frac{1}{3}$.

This gives us:

$$\frac{1}{3}S = \frac{1}{3} + \frac{2}{3^2} + \frac{7}{3^3} + \frac{12}{3^4} + \dots$$

Notice what happened? By multiplying by $\frac{1}{3}$, we have shifted every term one position to the right. Now, when we subtract $\frac{1}{3}S$ from $S$, the terms with the same denominators will align perfectly. This is the moment where the chaos turns into order.

### The Subtraction Magic

Let us perform the subtraction:

$$S - \frac{1}{3}S = 1 + \left(\frac{2}{3} - \frac{1}{3}ight) + \left(\frac{7}{3^2} - \frac{2}{3^2}ight) + \left(\frac{12}{3^3} - \frac{7}{3^3}ight) + \dots$$

On the left, we have $\frac{2}{3}S$. On the right, the first term $1$ remains untouched. The subsequent terms become:

$$\frac{2}{3}S = 1 + \frac{1}{3} + \frac{5}{3^2} + \frac{5}{3^3} + \frac{5}{3^4} + \dots$$

Do you see it? The constant difference of $5$ has emerged! We can now group the first two terms and factor out the $5$ from the rest:

$$\frac{2}{3}S = \left(1 + \frac{1}{3}ight) + 5 \left( \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \dots ight)$$

This simplifies to:

$$\frac{2}{3}S = \frac{4}{3} + 5 \left( \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \dots ight)$$

### The Final Convergence

The series inside the bracket is a standard infinite GP with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{3}$. Using the formula $S_{\infty} = \frac{a}{1-r}$, we get:

$$S_{gp} = \frac{1/9}{1 - 1/3} = \frac{1/9}{2/3} = \frac{1}{9} 	imes \frac{3}{2} = \frac{1}{6}$$

Substituting this back into our main equation, we get:

$$\frac{2}{3}S = \frac{4}{3} + 5 \left(\frac{1}{6}ight) = \frac{4}{3} + \frac{5}{6}$$

Finding a common denominator:

$$\frac{2}{3}S = \frac{8}{6} + \frac{5}{6} = \frac{13}{6}$$

Finally, multiplying by $\frac{3}{2}$, we find:

$$S = \frac{13}{6} 	imes \frac{3}{2} = \frac{13}{4}$$

And there it is! The infinite sum collapses into the beautiful, finite value of **$\frac{13}{4}$**. You have successfully tamed the infinite series!

Animated Solution for Mathematics - Sequence and Series: The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3 ^{2}} + \frac{12}{3 ^{3}} + \frac{17}{3 ^{4}} + \frac{22}{3 ^{5}} + \dots$ is equal to

Select Answer:

Visualized Solution

The Infinite Series

Identifying the Pattern

The AGP Method

Shifting and Subtracting

Performing the Subtraction

Grouping the Terms

The Infinite GP

Summing the GP

Computing the GP Sum

Back to the Main Equation

Solving for $\frac{2}{3} S$

Final Value of $S$

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

Analyzing the Setup

The Art of the Shift

The Subtraction Magic

The Final Convergence

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