Let $$\frac{1}{x_1}, \frac{1}{x_2}, \ldots, \frac{1}{x_n}$$ ($$x_i \neq 0$$ for i = 1, 2, ..., n) be in A.P. such that $$x_1 = 4$$ and $$x_{21} = 20$$. If n is the least positive integer for which $$x_n \gt 50$$, then $$\sum_{i=1}^{n}\left(\frac{1}{x_i}\right)$$ is equal to:

We are told that the quantities $$\dfrac1{x_1},\dfrac1{x_2},\ldots ,\dfrac1{x_n}$$ form an arithmetic progression (A.P.).

In any A.P., if the first term is denoted by $$a$$ and the common difference by $$d$$, then the $$k^{\text{th}}$$ term is given by the formula

$$a_k=a+(k-1)d.$$

Here the first term of the A.P. of reciprocals is

$$a=\dfrac1{x_1}=\dfrac14,$$

because $$x_1=4.$$

We are also given $$x_{21}=20,$$ therefore

$$a_{21}=\dfrac1{x_{21}}=\dfrac1{20}.$$

Using the general term formula for $$k=21$$ we write

$$\dfrac1{20}=a+(21-1)d.$$

Substituting $$a=\dfrac14$$ we obtain

$$\dfrac1{20}=\dfrac14+20d.$$

We isolate $$d$$ step by step:

$$20d=\dfrac1{20}-\dfrac14,$$

$$20d=\dfrac1{20}-\dfrac5{20}=-\dfrac4{20}=-\dfrac15,$$

$$d=\dfrac{-1/5}{20}=-\dfrac1{100}.$$

Thus the common difference of the reciprocals is negative: $$d=-\dfrac1{100}.$$ Because the reciprocals decrease by $$\dfrac1{100}$$ each step, the actual $$x_i$$ increase with $$i$$.

Next we must find the least positive integer $$n$$ for which $$x_n\gt 50.$$ The condition $$x_n\gt 50$$ is equivalent to

$$\dfrac1{x_n}\lt \dfrac1{50}.$$

Again using the general term formula for reciprocals we write

$$\dfrac1{x_n}=a+(n-1)d=\dfrac14+(n-1)\!\left(-\dfrac1{100}\right)=\dfrac14-\dfrac{\,n-1\,}{100}.$$

We demand

$$\dfrac14-\dfrac{\,n-1\,}{100}\lt \dfrac1{50}.$$

To compare the fractions conveniently we convert everything to the common denominator $$100$$:

$$\dfrac14=\dfrac{25}{100},\qquad \dfrac1{50}=\dfrac{2}{100}.$$

Hence the inequality becomes

$$\dfrac{25}{100}-\dfrac{\,n-1\,}{100}\lt \dfrac{2}{100}.$$

Multiplying by $$100$$ (a positive number, so the sense of the inequality is preserved) we get

$$25-(n-1)\lt 2.$$

Now simplify step by step:

$$25-n+1\lt 2,$$

$$26-n\lt 2,$$

$$-n\lt -24.$$

When we multiply both sides by $$-1$$ we must reverse the inequality sign:

$$n\gt 24.$$

The least positive integer satisfying this is clearly

$$n=25.$$

Therefore the required sum of reciprocals involves the first $$25$$ terms of the A.P.

For the sum of the first $$n$$ terms of an A.P. we repeatedly use the well-known formula

$$S_n=\frac{n}{2}\Bigl(2a+(n-1)d\Bigr),$$

or, equivalently, $$S_n=\dfrac{n}{2}(a_1+a_n).$$ Either form is correct; we shall employ the first.

Substituting $$n=25,\;a=\dfrac14,\;d=-\dfrac1{100}$$ we write

$$S_{25}=\frac{25}{2}\Bigl(2\cdot\dfrac14+(25-1)\!\left(-\dfrac1{100}\right)\Bigr).$$

Compute each part carefully. First,

$$2\cdot\dfrac14=\dfrac12.$$

Next,

$$(25-1)d=24\left(-\dfrac1{100}\right)=-\dfrac{24}{100}=-\dfrac6{25}.$$

Adding these two contributions yields

$$2a+(n-1)d=\dfrac12-\dfrac6{25}.$$

To combine the fractions, convert $$\dfrac12$$ to the denominator $$100$$ for convenience:

$$\dfrac12=\dfrac{50}{100},\qquad \dfrac6{25}=\dfrac{24}{100}.$$

Hence

$$\dfrac12-\dfrac6{25}=\dfrac{50}{100}-\dfrac{24}{100}=\dfrac{26}{100}=\dfrac{13}{50}.$$

We can now complete the sum:

$$S_{25}= \frac{25}{2}\left(\dfrac{13}{50}\right)=\dfrac{25\times13}{2\times50}=\dfrac{325}{100}=\dfrac{13}{4}.$$

Thus the value of $$\displaystyle\sum_{i=1}^{n}\left(\dfrac1{x_i}\right)$$ is $$\dfrac{13}{4}.$$

Hence, the correct answer is Option C.