The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ......, is :

We have the decimal sequence $$0.7,\, 0.77,\, 0.777,\,\dots$$ and we must add its first twenty terms.

Let $$T_n$$ be the $$n^{\text{th}}$$ term, i.e. $$T_n = 0.\underbrace{77\dots7}_{n\ \text{digits of}\ 7}.$$

To convert such a terminating decimal into a fraction, observe that each $$7$$ can be factored out of the digits:

$$T_n = 7\left(0.\underbrace{11\dots1}_{n\ \text{digits of}\ 1}\right).$$

The block $$0.111\dots1$$ with exactly $$n$$ digits is a finite geometric series:

$$0.111\dots1 = 10^{-1} + 10^{-2} + \dots + 10^{-n}.$$

For a geometric series with first term $$a = 10^{-1}$$ and common ratio $$r = 10^{-1},$$ the sum of the first $$n$$ terms is

$$S_n = a\,\frac{1-r^{\,n}}{1-r}.$$

Substituting $$a=10^{-1}$$ and $$r=10^{-1},$$ we get

$$0.111\dots1 = 10^{-1}\,\frac{1-(10^{-1})^{n}}{1-10^{-1}} = 10^{-1}\,\frac{1-10^{-n}}{\tfrac{9}{10}} = \frac{1}{9}\,(1-10^{-n}).$$

Hence

$$T_n = 7\left[\frac{1}{9}\,(1-10^{-n})\right] = \frac{7}{9}\,(1-10^{-n}).$$

Now we add these terms from $$n=1$$ to $$n=20$$. Denote the required sum by $$S_{20}$$:

$$S_{20} = \sum_{n=1}^{20} T_n = \frac{7}{9}\,\sum_{n=1}^{20}\,(1-10^{-n}).$$

Split the summation into two simpler sums:

$$S_{20} = \frac{7}{9}\left[\sum_{n=1}^{20} 1\;-\;\sum_{n=1}^{20}10^{-n}\right].$$

The first sum is just $$20,$$ because the term $$1$$ is repeated twenty times.

The second sum is another geometric series with first term $$10^{-1}$$ and common ratio $$10^{-1}.$$ Using the same formula again, its value is

$$\sum_{n=1}^{20}10^{-n} = 10^{-1}\,\frac{1-(10^{-1})^{20}}{1-10^{-1}} = 10^{-1}\,\frac{1-10^{-20}}{\tfrac{9}{10}} = \frac{1}{9}\,(1-10^{-20}).$$

So

$$S_{20} = \frac{7}{9}\left[20 - \frac{1}{9}\,(1-10^{-20})\right].$$

Simplify the bracketed expression step by step. First handle the constant part:

$$20 - \frac{1}{9} = \frac{180}{9} - \frac{1}{9} = \frac{179}{9}.$$

Now keep the $$10^{-20}$$ term separate:

$$20 - \frac{1}{9}(1-10^{-20}) = \frac{179}{9} + \frac{10^{-20}}{9}.$$ Thus

$$S_{20} = \frac{7}{9}\left(\frac{179 + 10^{-20}}{9}\right) = \frac{7\,(179 + 10^{-20})}{81}.$$

This matches option A.

Hence, the correct answer is Option 1.