The sum of 10 terms of the series `0.7+ 0.77+ 0.777+………`is -

Answer

To find the sum of the first 10 terms of the series \(0.7 + 0.77 + 0.777 + \ldots\), we can express the series in a more manageable form.

### Step 1: Express the terms in a more usable format
The terms of the series can be rewritten as:
- 1st term: \(0.7 = \frac{7}{10}\)
- 2nd term: \(0.77 = \frac{77}{100} = \frac{7 \times 11}{100}\)
- 3rd term: \(0.777 = \frac{777}{1000} = \frac{7 \times 111}{1000}\)

We can see that the \(n\)-th term can be expressed as:
\[ T_n = \frac{7 \times (10^n - 1)/9}{10^n} = \frac{7}{9} \left(1 - \frac{1}{10^n}\right) \]

### Step 2: Find the sum of the first 10 terms
The sum \(S_n\) of the first \(n\) terms can be calculated as:
\[ S_n = \sum_{k=1}^{n} T_k = \frac{7}{9} \sum_{k=1}^{n} \left(1 - \frac{1}{10^k}\right) \]

### Step 3: Calculate the sum of the series
We can separate the sum:
\[ S_n = \frac{7}{9} \left( \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{1}{10^k} \right) \]
The first sum is simply \(n\):
\[ \sum_{k=1}^{n} 1 = n \]

The second sum is a geometric series:
\[ \sum_{k=1}^{n} \frac{1}{10^k} = \frac{\frac{1}{10} \left(1 - \left(\frac{1}{10}\right)^n\right)}{1 - \frac{1}{10}} = \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \]

### Step 4: Substitute back into the sum formula
Now substituting back:
\[ S_n = \frac{7}{9} \left(n - \frac{1}{9} \left(1 - \frac{1}{10^n}\right)\right) \]

### Step 5: Calculate \(S_{10}\)
Now, substituting \(n = 10\):
\[ S_{10} = \frac{7}{9} \left(10 - \frac{1}{9} \left(1 - \frac{1}{10^{10}}\right)\right) \]
\[ = \frac{7}{9} \left(10 - \frac{1}{9} + \frac{1}{9 \times 10^{10}}\right) \]
\[ = \frac{7}{9} \left(\frac{90 - 1}{9} + \frac{1}{9 \times 10^{10}}\right) \]
\[ = \frac{7}{9} \left(\frac{89}{9} + \frac{1}{9 \times 10^{10}}\right) \]
\[ = \frac{7 \times 89}{81} + \frac{7}{81 \times 10^{10}} \]

### Final Calculation
Calculating \(S_{10}\):
\[ S_{10} = \frac{623}{81} + \frac{7}{81 \times 10^{10}} \]

Thus, the sum of the first 10 terms of the series is approximately:
\[ S_{10} \approx 7.6864 \text{ (ignoring the very small term)} \]