The third term of a geometric progression is -108 and the sixth term is 32. Find i) The common ratio and the first term.

ii) Geometric sequence

iii) 10th term

Given that the third term of a geometric progression (GP) is \(-108\) and the sixth term is \(32\), we are to find the common ratio, the first term, and other related details.

### Step 1: Find the Common Ratio (r)

Let's denote the first term by \(a\) and the common ratio by \(r\). The \(n\)th term of a geometric sequence can be expressed as:

\[
T_n = ar^{n-1}
\]

For the third term (\(T_3\)):

\[
T_3 = ar^{3-1} = ar^2 = -108
\]

For the sixth term (\(T_6\)):

\[
T_6 = ar^{6-1} = ar^5 = 32
\]

We now have two equations:

\[
ar^2 = -108 \quad \text{(Equation 1)}
\]
\[
ar^5 = 32 \quad \text{(Equation 2)}
\]

### Step 2: Solve for the Common Ratio (r)

Divide Equation 2 by Equation 1:

\[
\frac{ar^5}{ar^2} = \frac{32}{-108}
\]

Simplifying:

\[
r^3 = \frac{32}{-108} = \frac{32}{-108} = -\frac{8}{27}
\]

Thus,

\[
r = \sqrt[3]{-\frac{8}{27}} = -\frac{2}{3}
\]

So, the common ratio \(r\) is \( \boxed{-\frac{2}{3}} \).

### Step 3: Find the First Term (a)

Using the value of \( r \) in Equation 1:

\[
ar^2 = -108
\]

Substitute \( r = -\frac{2}{3} \):

\[
a \left(-\frac{2}{3}\right)^2 = -108
\]

\[
a \cdot \frac{4}{9} = -108
\]

Multiply both sides by \( \frac{9}{4} \):

\[
a = -108 \times \frac{9}{4} = -27 \times 9 = -243
\]

So, the first term \( a \) is \( \boxed{-243} \).

### Step 4: Write the Geometric Sequence

With \( a = -243 \) and \( r = -\frac{2}{3} \), the first few terms of the geometric sequence are:

1. First term: \( T_1 = -243 \)
2. Second term: \( T_2 = -243 \times \left(-\frac{2}{3}\right) = 162 \)
3. Third term: \( T_3 = 162 \times \left(-\frac{2}{3}\right) = -108 \) 
4. Fourth term: \( T_4 = -108 \times \left(-\frac{2}{3}\right) = 72 \)
5. Fifth term: \( T_5 = 72 \times \left(-\frac{2}{3}\right) = -48 \)
6. Sixth term: \( T_6 = -48 \times \left(-\frac{2}{3}\right) = 32 \)

So, the geometric sequence is:

\[
\boxed{-243, 162, -108, 72, -48, 32, \dots}
\]

### Step 5: Find the 10th Term

To find the 10th term (\(T_{10}\)):

\[
T_{10} = ar^{10-1} = ar^9
\]

Substitute \( a = -243 \) and \( r = -\frac{2}{3} \):

\[
T_{10} = -243 \times \left(-\frac{2}{3}\right)^9
\]

Since \(\left(-\frac{2}{3}\right)^9\) is negative, we get:

\[
T_{10} = -243 \times -\frac{512}{19683} = \frac{243 \times 512}{19683} = \frac{124416}{19683} = \boxed{6.32} \quad \text{(approx)}
\]

Would you like any further details or have any questions? Here are some related questions:

1. What is the general formula for the \( n \)th term of a geometric sequence?
2. How do you find the sum of the first \( n \) terms in a geometric series?
3. How would the sequence change if the common ratio was a positive number?
4. What happens if the common ratio is a fraction greater than 1?
5. How can you determine if a sequence is geometric based on the terms provided?

**Tip:** When working with geometric sequences, it's important to consider whether the common ratio is positive or negative, as this will affect the sign of the terms in the sequence.

Learn how to find the common ratio and first term of a geometric progression (GP) given specific terms. Solve for the 10th term and understand the sequence progression. Suitable for Grades 9-12.

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