The nth term of a geometric progression (G.P.) is given by T_n = 5 imes 22-n. Write down the first four terms and calculate the sum of the first six terms.

$$T_4 = 5 \times 2^{2-4} = 5 \times 2^{-2} = 5 \times \frac{1}{4} = \frac{5}{4}$$

The first four terms are: $\boxed{10, 5, \frac{5}{2}, \frac{5}{4}}$

From part (a), the first term is $a = T_1 = 10$.
The common ratio $r$ is found by dividing any term by its preceding term:

$$r = \frac{T_2}{T_1} = \frac{5}{10} = \frac{1}{2}$$

The formula for the sum of the first $n$ terms is $S_n = \frac{a(1 - r^n)}{1 - r}$.
We need to find the sum of the first six terms, so $n=6$, $a=10$, and $r=\frac{1}{2}$.

$$S_6 = \frac{10 \left(1 - \left(\frac{1}{2}\right)^6\right)}{1 - \frac{1}{2}}$$

$$S_6 = \frac{10 \left(1 - \frac{1}{64}\right)}{\frac{1}{2}}$$ $$S_6 = \frac{10 \left(\frac{64}{64} - \frac{1}{64}\right)}{\frac{1}{2}}$$ $$S_6 = \frac{10 \left(\frac{63}{64}\right)}{\frac{1}{2}}$$ $$S_6 = 10 \times \frac{63}{64} \times 2$$ $$S_6 = \frac{20 \times 63}{64}$$ $$S_6 = \frac{5 \times 63}{16}$$ $$S_6 = \frac{315}{16}$$

The sum of the first six terms is: $\boxed{\frac{315}{16}}$