The difference between the fourth term and the first term of a Geometrical Progression is 52. If the sum of its first three terms is 26, then the sum of the first six terms of the progression is

Let the first term of the G.P. be $$a$$ and the common ratio be $$r$$.
The terms are $$a,\,ar,\,ar^2,\,ar^3,\ldots$$

Given data
Fourth-term minus first-term: $$ar^3 - a = 52$$ $$-(1)$$
Sum of first three terms: $$a + ar + ar^2 = 26$$ $$-(2)$$

Factor out $$a$$ in both equations.

From $$-(1):\; a(r^3 - 1) = 52$$

From $$-(2):\; a(1 + r + r^2) = 26$$

Divide $$-(1)$$ by $$-(2)$$ to eliminate $$a$$:

$$\frac{a(r^3-1)}{a(1+r+r^2)}=\frac{52}{26}=2$$

$$\Rightarrow\; \frac{r^3-1}{1+r+r^2}=2$$

Use the identity $$r^3-1=(r-1)(r^2+r+1).$$
Thus, $$\frac{(r-1)(r^2+r+1)}{1+r+r^2}=r-1.$$

So $$r-1 = 2 \;\Rightarrow\; r = 3.$$

Substitute $$r=3$$ in $$-(2):$$

$$a(1+3+9)=26 \;\Longrightarrow\; a \times 13 = 26 \;\Longrightarrow\; a = 2.$$

Sum of the first six terms
For a G.P., $$S_6 = a\frac{r^6-1}{r-1}.$$ Substitute $$a = 2$$ and $$r = 3$$:

$$S_6 = 2 \cdot \frac{3^6 - 1}{3 - 1} = 2 \cdot \frac{729 - 1}{2} = 2 \cdot \frac{728}{2} = 728.$$

Hence, the sum of the first six terms is 728.
Option C which is: 728