MATHEMATICS DEPARTMENT HOLHDAY ASSIGNMENT GRADE 11 Due date: first week of reporting $$ \left.s(-1)^{-1}\right)-2 $$ 1. The first 3 terms of a geometric progression are $$ 3 a-2,3 a+6 $$ and $$ 3 a+30 $$. Find the 5. $$ T_{1}=\frac{T_{1}}{T_{2}}=\frac{T_{2}}{T_{3}} $$ a) Value of a b) $$ \mathbf{N}^{\text {th }} $$ term c) sum of the first 4 terms $$ \frac{3 a-2}{3 a+6} \times \frac{3 a+6}{3 a+30} $$

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Certainly! Let's solve the given problems step by step.

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### **Problem Statement:**

The first 3 terms of a geometric progression (GP) are:
- $$ T_1 = 3a - 2 $$
- $$ T_2 = 3a + 6 $$
- $$ T_3 = 3a + 30 $$

Given that it's a GP, the ratio between consecutive terms is constant. You'll be required to:

**a) Find the value of $$ a $$.**

**b) Determine the $$ \mathbf{N}^{\text{th}} $$ term of the GP.**

**c) Find the sum of the first 4 terms of the GP.**

---

### **a) Finding the Value of $$ a $$**

**Step 1: Understanding the Common Ratio**

In a GP, the ratio of the second term to the first term is equal to the ratio of the third term to the second term. Mathematically,

$$
\frac{T_2}{T_1} = \frac{T_3}{T_2}
$$

**Step 2: Setting Up the Equation**

Substitute the given terms:

$$
\frac{3a + 6}{3a - 2} = \frac{3a + 30}{3a + 6}
$$

**Step 3: Cross-Multiplying to Solve for $$ a $$**

$$
(3a + 6)^2 = (3a - 2)(3a + 30)
$$

Expand both sides:

$$
9a^2 + 36a + 36 = 9a^2 + 84a - 60
$$

Subtract $$ 9a^2 $$ from both sides:

$$
36a + 36 = 84a - 60
$$

Bring like terms together:

$$
36 + 60 = 84a - 36a
$$

$$
96 = 48a
$$

Divide both sides by 48:

$$
a = \frac{96}{48} = 2
$$

**Answer:**  
**a) The value of $$ a $$ is $$ 2 $$.**

---

### **b) Determining the $$ \mathbf{N}^{\text{th}} $$ Term of the GP**

**Step 1: Calculate the First Three Terms Using $$ a = 2 $$**

$$
T_1 = 3(2) - 2 = 6 - 2 = 4
$$
$$
T_2 = 3(2) + 6 = 6 + 6 = 12
$$
$$
T_3 = 3(2) + 30 = 6 + 30 = 36
$$

**Step 2: Find the Common Ratio ($$ r $$)**

$$
r = \frac{T_2}{T_1} = \frac{12}{4} = 3
$$

**Step 3: Use the Formula for the $$ N^{\text{th}} $$ Term**

The general formula for the $$ N^{\text{th}} $$ term in a GP is:

$$
T_n = T_1 \times r^{(n-1)}
$$

Substitute the known values:

$$
T_n = 4 \times 3^{(n-1)}
$$

**Answer:**  
**b) The $$ N^{\text{th}} $$ term of the GP is $$ T_n = 4 \times 3^{n-1} $$.**

---

### **c) Finding the Sum of the First 4 Terms of the GP**

**Step 1: Use the Sum Formula for a GP**

The sum of the first $$ n $$ terms of a GP is given by:

$$
S_n = T_1 \times \frac{r^n - 1}{r - 1}
$$

**Step 2: Substitute the Known Values**

For $$ n = 4 $$:

$$
S_4 = 4 \times \frac{3^4 - 1}{3 - 1} = 4 \times \frac{81 - 1}{2} = 4 \times \frac{80}{2} = 4 \times 40 = 160
$$

**Alternatively, you can list out the first four terms and sum them:**

$$
T_1 = 4, \quad T_2 = 12, \quad T_3 = 36, \quad T_4 = 108
$$
$$
S_4 = 4 + 12 + 36 + 108 = 160
$$

**Answer:**  
**c) The sum of the first 4 terms of the GP is $$ 160 $$.**

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### **Summary of Answers:**

a) **$$ a = 2 $$**  
b) **$$ T_n = 4 \times 3^{n-1} $$**  
c) **Sum of the first 4 terms is $$ 160 $$**
a) $$ a = 2 $$ b) $$ T_n = 4 \times 3^{n-1} $$ c) Sum of the first 4 terms is 160

MATHEMATICS DEPARTMENT
HOLHDAY ASSIGNMENT GRADE 11
Due date: first week of reporting

The first 3 terms of a geometric progression are and . Find the

a) Value of a
b) term
c) sum of the first 4 terms

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MATHEMATICS DEPARTMENT HOLHDAY ASSIGNMENT GRADE 11 Due date: first week of reporting The first 3 terms of a geometric progression are and . Find the a) Value of a b) term c) sum of the first 4 terms