The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ... is (A) 7/81 (179 - 10⁻²⁰) (B) 7/9 (9

Q: Wait, why do we multiply and divide by $9$? Is there a specific reason for that?

That is a fantastic question! We multiply by $9$ because it transforms the sequence into a form that is easily decomposable. Notice that $0.9 = 1 - 0.1$, $0.99 = 1 - 0.01$, and so on. By creating these nines, we force the sequence to become a difference between a constant ($1$) and a power of $10$. It is a classic technique for handling repeating decimal patterns in JEE problems.

Q: Why can't we just use the formula for the sum of a GP directly on the original sequence?

Because the original sequence $0.7, 0.77, 0.777, \dots$ is not a GP! In a GP, the ratio between consecutive terms must be constant. Here, $\frac{0.77}{0.7} = 1.1$, but $\frac{0.777}{0.77} \approx 1.009$. Since the ratio changes, the standard GP formula doesn't apply until we perform the algebraic manipulation to reveal the underlying structure.

Q: Why does the negative sign become positive at the end of the calculation?

Keep a close eye on the brackets! We have $S = \frac{7}{9} [ 20 - (\text{Sum of GP}) ]$. When you substitute the GP sum formula, you get a term like $(1 - 10^{-20})$. When you distribute the negative sign from the original expression into this result, the $-1$ becomes negative, and the $-(-10^{-20})$ becomes positive. It is a common place to lose a sign, so always expand your brackets carefully.

Question

The sum of first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

Options:

(A) $\frac{7}{81}(179 - 10^{-20})$

(B) $\frac{7}{9}(99 - 10^{-20})$

(C) $\frac{7}{81}(179 + 10^{-20})$

(D) $\frac{7}{9}(99 + 10^{-20})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$.

At first glance, you might be tempted to look for a common difference or a common ratio, but you will quickly realize that neither exists. This is not an Arithmetic Progression, nor is it a Geometric Progression.

It is a hybrid, a pattern that requires us to peel back the layers using the art of algebraic manipulation.

### The Magic of the Nine

To solve this, we need to transform the sequence into something our mathematical toolkit can handle. The educator's secret weapon here is the $9/9$ trick.

Why nine? Because nine is just one step away from ten, and powers of ten are the language of decimals. Let us factor out the $7$ from each term:

$$S = 7 [ 0.1 + 0.11 + 0.111 + \dots ]$$

Now, we multiply and divide by $9$ to get:

$$S = \frac{7}{9} [ 0.9 + 0.99 + 0.999 + \dots ]$$

Suddenly, the sequence starts to look familiar. We have turned a messy decimal sequence into a series of numbers that are almost integers.

### The Decomposition: Breaking the Pattern

This is the pivotal moment in our journey. We can rewrite each term $0.9, 0.99, 0.999, \dots$ as a difference: $0.9 = 1 - 0.1$, $0.99 = 1 - 0.01$, and $0.999 = 1 - 0.001$.

In the language of powers, this is $(1 - 10^{-1}), (1 - 10^{-2}), (1 - 10^{-3}), \dots, (1 - 10^{-20})$. By substituting this back into our sum, we get:

$$S = \frac{7}{9} [ (1 - 10^{-1}) + (1 - 10^{-2}) + \dots + (1 - 10^{-20}) ]$$

Now, we can separate the constants from the powers of ten. We have $20$ ones, which sum to $20$, and a series of negative powers of ten:

$$S = \frac{7}{9} [ 20 - (10^{-1} + 10^{-2} + \dots + 10^{-20}) ]$$

### The Geometric Progression Revealed

Now, look at the second part of the expression: $(10^{-1} + 10^{-2} + \dots + 10^{-20})$. This is a classic Geometric Progression (GP) where the first term $a = 10^{-1} = 0.1$ and the common ratio $r = 10^{-1} = 0.1$.

The number of terms is $n = 20$. Using the sum formula for a GP, $S_n = \frac{a(1-r^n)}{1-r}$, we calculate:

$$	ext{Sum} = \frac{0.1(1 - 0.1^{20})}{1 - 0.1} = \frac{0.1}{0.9}(1 - 10^{-20}) = \frac{1}{9}(1 - 10^{-20})$$

### The Final Synthesis

We are almost at the finish line. Let us plug this back into our main equation:

$$S = \frac{7}{9} [ 20 - \frac{1}{9}(1 - 10^{-20}) ]$$

To simplify this, we take $\frac{1}{9}$ as a common factor from inside the bracket:

$$S = \frac{7}{9} \cdot \frac{1}{9} [ 180 - (1 - 10^{-20}) ]$$

This gives us:

$$S = \frac{7}{81} [ 180 - 1 + 10^{-20} ]$$

Simplifying the constants, we arrive at the final result:

**$$S = \frac{7}{81} [ 179 + 10^{-20} ]$$**

This elegant result matches our option perfectly. Remember, in JEE Advanced, the complexity is often just a veil. Once you apply the right transformation, the problem collapses into a simple, beautiful solution.

Animated Solution for Mathematics - Sequence and Series: The sum of first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

Select Answer:

Visualized Solution

Identify the Sequence

Factor out the Common Digit

The Standard Multiplier Trick

Rewrite as Powers of Ten

Separate the Sums

Sum of the Constant Part

Identify the G.P. Parameters

Apply G.P. Sum Formula

Simplify the G.P. Sum

Combine and Factorize

Final Result

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

Analyzing the Setup

The Magic of the Nine

The Decomposition

The Geometric Progression Revealed

The Final Synthesis

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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; } The sum of first 20 terms of the sequence 0.7,0.77,0.777,… is