Here are the solutions to the questions: 4. Find the ninth term and the general term of the progression $\frac{1}{4}, -\frac{1}{2}, 1, -2$. Step 1: 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). Step 2: Find the general term ($T_n$). The formula for the general term of a GP is $T_n = a \cdot r^{n-1}$. Substitute $a = \frac{1}{4}$ and $r = -2$: $$T_n = \frac{1}{4} (-2)^{n-1}$$ Step 3: Find the ninth term ($T_9$). Substitute $n=9$ into the general term formula: $$T_9 = \frac{1}{4} (-2)^{9-1}$$ $$T_9 = \frac{1}{4} (-2)^8$$ Calculate $(-2)^8$: $$(-2)^8 = 256$$ $$T_9 = \frac{1}{4} \times 256$$ $$T_9 = 64$$ 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 $r=2$. Step 1: Use the formula for the sum of the first $n$ terms of a GP. 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 $n=8$. Step 2: Substitute the values into the formula. $$S_8 = \frac{5(2^8 - 1)}{2 - 1}$$ $$S_8 = \frac{5(256 - 1)}{1}$$ $$S_8 = 5(255)$$ $$S_8 = 1275$$ 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. Step 1: Write down the given information using the GP term formula $T_n = a \cdot r^{n-1}$. $T_1 = a = 153$ $T_6 = a \cdot r^{6-1} = a \cdot r^5 = \frac{17}{27}$ Step 2: Find the common ratio ($r$). Substitute $a=153$ into the equation for $T_6$: $$153 \cdot r^5 = \frac{17}{27}$$ $$r^5 = \frac{17}{27 \times 153}$$ Simplify the fraction: $153 = 9 \times 17$. $$r^5 = \frac{17}{27 \times 9 \times 17}$$ $$r^5 = \frac{1}{27 \times 9}$$ $$r^5 = \frac{1}{3^3 \times 3^2}$$ $$r^5 = \frac{1}{3^5}$$ Therefore, $r = \frac{1}{3}$. Step 3: Find the sum of the first four terms ($S_4$). 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 $n=4$. $$S_4 = \frac{153 \left(1 - \left(\frac{1}{3}\right)^4\right)}{1 - \frac{1}{3}}$$ Calculate $\left(\frac{1}{3}\right)^4$: $$\left(\frac{1}{3}\right)^4 = \frac{1}{81}$$ Calculate $1 - \frac{1}{3}$: $$1 - \frac{1}{3} = \frac{2}{3}$$ Substitute these values back into the formula: $$S_4 = \frac{153 \left(1 - \frac{1}{81}\right)}{\frac{2}{3}}$$ $$S_4 = \frac{153 \left(\frac{81 - 1}{81}\right)}{\frac{2}{3}}$$ $$S_4 = \frac{153 \left(\frac{80}{81}\right)}{\frac{2}{3}}$$ $$S_4 = 153 \times \frac{80}{81} \times \frac{3}{2}$$ Simplify the expression: $$S_4 = \frac{153 \times 80 \times 3}{81 \times 2}$$ $$S_4 = \frac{(9 \times 17) \times (2 \times 40) \times 3}{(9 \times 9) \times 2}$$ $$S_4 = \frac{17 \times 40 \times 3}{9}$$ $$S_4 = \frac{17 \times 40 \times 1}{3}$$ $$S_4 = \frac{680}{3}$$ 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. Step 1: Write down the given information using the GP term formula $T_n = a \cdot r^{n-1}$. $T_2 = a \cdot r^{2-1} = a \cdot r = 35$ (Equation 1) $T_4 = a \cdot r^{4-1} = a \cdot r^3 = 875$ (Equation 2) Step 2: Find the common ratio ($r$). Divide Equation 2 by Equation 1: $$\frac{a \cdot r^3}{a \cdot r} = \frac{875}{35}$$ $$r^2 = 25$$ $$r = \pm \sqrt{25}$$ $$r = \pm 5$$ Step 3: Find the first term ($a$) and the fifth term ($T_5$) for each possible value of $r$. 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: $r=-5$