In a geometric progression we have: t5=8, tn=0.5 and n=9. Determine S9.

Given a geometric progression with t_5 = 8 , t_n = 0.5 , and n = 9 . We want to find the sum of the first nine terms.

The formula for the general term of a geometric progression is given by:
t_n = a \cdot r^{n-1},
where:
t_n is the n^{th} term,
a is the first term,
r is the common ratio,
and n is the number of terms.

We are given that t_5 = 8 and t_n = 0.5 .
For t_5 = 8 , we have:
t_5 = a \cdot r^{5-1} = 8.
a \cdot r^4 = 8. \qquad (1)

For t_9 = 0.5 , we have:
t_9 = a \cdot r^{9-1} = 0.5.
a \cdot r^8 = 0.5. \qquad (2)

Dividing equation (2) by equation (1):
\frac{a \cdot r^8}{a \cdot r^4} = \frac{0.5}{8}.
r^4 = \frac{0.5}{8} = \frac{1}{16}.
r = \sqrt[4]{\frac{1}{16}} = \frac{1}{2}.

Now, substitute r = \frac{1}{2} into equation (1):
a \cdot (\frac{1}{2})^4 = 8.
a \cdot \frac{1}{16} = 8.
a = 8 \cdot 16 = 128.

The sum of the first nine terms of the geometric progression is given by:
S_n=\frac{a\cdot(1-r^n)}{1-r}.
Substitute a = 128 , r = \frac{1}{2} , and n = 9 into the formula:
S_9=\frac{128\cdot(1-\frac{1}{2}^9)}{1-\frac{1}{2}}.

S_9=255.5