Identify the first term (a) and the common ratio (r).

The first term is $a = \frac{1}{4}$.
The common ratio $r$ is found by dividing any term by its preceding term:
$r = \frac{-\frac{1}{2}}{\frac{1}{4}} = -\frac{1}{2} \times 4 = -2$.
Check with the next terms: $\frac{1}{-\frac{1}{2}} = -2$, $\frac{-2}{1} = -2$.
This is a geometric progression (GP).

The formula for the general term of a GP is $T_n = a \cdot r^{n-1}$.
Substitute $a = \frac{1}{4}$ and $r = -2$:

Substitute $n=9$ into the general term formula:

Calculate $(-2)^8$:

The general term is $\boxed{T_n = \frac{1}{4}(-2)^{n-1}}$ and the ninth term is $\boxed{64}$.

5. Find the sum of the first 8 terms of the G.P with $a=5$ and $r=2$.

The formula for the sum of the first $n$ terms of a GP when $r \neq 1$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.
Given $a=5$, $r=2$, and $n=8$.

The sum of the first 8 terms is $\boxed{1275}$.

6. The $1^{st}$ and $6^{th}$ terms of a G.P are 153 and $17/27$. Find the sum of the first four terms.

$T_1 = a = 153$
$T_6 = a \cdot r^{6-1} = a \cdot r^5 = \frac{17}{27}$

Substitute $a=153$ into the equation for $T_6$:

Simplify the fraction: $153 = 9 \times 17$.

Therefore, $r = \frac{1}{3}$.

The formula for the sum of the first $n$ terms of a GP when $|r| < 1$ is $S_n = \frac{a(1 - r^n)}{1 - r}$.
Given $a=153$, $r=\frac{1}{3}$, and $n=4$.

Calculate $\left(\frac{1}{3}\right)^4$:

Calculate $1 - \frac{1}{3}$:

Substitute these values back into the formula:

Simplify the expression:

The sum of the first four terms is $\boxed{\frac{680}{3}}$.

7. If $T_2=35$ and $T_4=875$ respectively. Find the first and fifth terms.

$T_2 = a \cdot r^{2-1} = a \cdot r = 35$ (Equation 1)
$T_4 = a \cdot r^{4-1} = a \cdot r^3 = 875$ (Equation 2)

Divide Equation 2 by Equation 1:

• Case 1: $r=5$ Substitute $r=5$ into Equation 1:

  $$a \cdot 5 = 35$$ $$a = \frac{35}{5}$$ $$a = 7$$

  The first term is $T_1 = a = 7$. The fifth term is $T_5 = a \cdot r^{5-1} = a \cdot r^4$:

  $$T_5 = 7 \cdot (5)^4$$ $$T_5 = 7 \cdot 625$$ $$T_5 = 4375$$

• Case 2: $r=-5$