Let’s work through each problem on the sheet step-by-step. I’ll start with problem 1 and provide a solution.

Problem 1: Finding the Sum of the First 10 Terms for each G.P.

Given:

1. $(i) \, 4, 2, 1, \frac{1}{2}, \dots$
2. $(ii) \, -2, 4, -8, 16, \dots$

Solution:

For a geometric progression (G.P.), the sum of the first $n$ terms $S_n$ is given by: $S_n = a \frac{1 - r^n}{1 - r} \quad \text{if } r \neq 1$ where:

• $a$ is the first term,
• $r$ is the common ratio,
• $n$ is the number of terms.

Part (i): $4, 2, 1, \frac{1}{2}, \dots$

• First term $a = 4$.
• Common ratio $r = \frac{2}{4} = \frac{1}{2}$.
• Number of terms $n = 10$.

Using the sum formula: $S_{10} = 4 \frac{1 - \left(\frac{1}{2}\right)^{10}}{1 - \frac{1}{2}}$ Calculate this to find the sum.

Part (ii): $-2, 4, -8, 16, \dots$

• First term $a = -2$.
• Common ratio $r = \frac{4}{-2} = -2$.
• Number of terms $n = 10$.

Using the sum formula: $S_{10} = -2 \frac{1 - (-2)^{10}}{1 - (-2)}$ Calculate this as well.

Would you like the complete calculations for each part, or should I proceed with other questions?

5 Related Questions:

1. How do you calculate the sum for an infinite geometric series?
2. What happens if the common ratio $r$ is negative?
3. How do you find the common ratio if only some terms are given?
4. How would you solve for $n$ if you know the sum but not the number of terms?
5. How does the sum formula change if the common ratio $r$ is equal to 1?

Tip: Remember that the sum formula for a G.P. only works if $r \neq 1$. If $r = 1$, the series is arithmetic, not geometric.