Find the sum of the indicated number of terms of each of the following geometric progressions:
a. (i) `sqrt(7),sqrt(21),3sqrt(7)`,…………..n terms.
(ii) `2,-1/2,1/8`, …………….n terms 12 terms
(iii)`1,1/3,1/9,`………………..,n terms 5 terms
b. (i) `x^(3),x^(5),x^(7)`.......,n terms `(x!= +-1)`
(ii) `1,-a,a^(2),-a^(3)`........, n terms `(a!= -1)`
(iii) `x^(2)-y^(2),x-y,(x-y)/(x+y)` ................ , n terms `(x+y!=1)`.

Answer

To find the sum of the indicated number of terms of each of the geometric progressions (GPs), we will use the formula for the sum of the first n terms of a GP:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{(if } r \neq 1\text{)} \]

Where:
- \( S_n \) is the sum of the first n terms,
- \( a \) is the first term,
- \( r \) is the common ratio,
- \( n \) is the number of terms.

Now, let's solve each part step by step.

### Part a

#### (i) \( \sqrt{7}, \sqrt{21}, 3\sqrt{7}, \ldots \) (n terms)

1. **Identify \( a \) and \( r \)**:
- \( a = \sqrt{7} \)
- To find \( r \), we calculate \( r = \frac{T_2}{T_1} = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} \).

2. **Apply the formula**:
\[ S_n = \frac{\sqrt{7}(1 - (\sqrt{3})^n)}{1 - \sqrt{3}} \]

3. **Rationalize the denominator**:
\[ S_n = \frac{\sqrt{7}(1 - (\sqrt{3})^n)}{1 - \sqrt{3}} \cdot \frac{1 + \sqrt{3}}{1 + \sqrt{3}} = \frac{\sqrt{7}(1 - (\sqrt{3})^n)(1 + \sqrt{3})}{(1 - 3)} = \frac{\sqrt{7}(1 - (\sqrt{3})^n)(1 + \sqrt{3})}{-2} \]

4. **Final expression**:
\[ S_n = -\frac{\sqrt{7}(1 - (\sqrt{3})^n)(1 + \sqrt{3})}{2} \]

#### (ii) \( 2, -\frac{1}{2}, \frac{1}{8}, \ldots \) (12 terms)

1. **Identify \( a \) and \( r \)**:
- \( a = 2 \)
- \( r = \frac{-\frac{1}{2}}{2} = -\frac{1}{4} \)

2. **Apply the formula**:
\[ S_{12} = \frac{2(1 - (-\frac{1}{4})^{12})}{1 - (-\frac{1}{4})} \]

3. **Simplify**:
\[ S_{12} = \frac{2(1 - \frac{1}{4^{12}})}{1 + \frac{1}{4}} = \frac{2(1 - \frac{1}{4^{12}})}{\frac{5}{4}} = \frac{8(1 - \frac{1}{4^{12}})}{5} \]

#### (iii) \( 1, \frac{1}{3}, \frac{1}{9}, \ldots \) (5 terms)

1. **Identify \( a \) and \( r \)**:
- \( a = 1 \)
- \( r = \frac{\frac{1}{3}}{1} = \frac{1}{3} \)

2. **Apply the formula**:
\[ S_5 = \frac{1(1 - (\frac{1}{3})^5)}{1 - \frac{1}{3}} = \frac{1(1 - \frac{1}{243})}{\frac{2}{3}} = \frac{3(1 - \frac{1}{243})}{2} \]

3. **Final expression**:
\[ S_5 = \frac{3 \cdot \frac{242}{243}}{2} = \frac{726}{486} = \frac{121}{81} \]

### Part b

#### (i) \( x^3, x^5, x^7, \ldots \) (n terms)

1. **Identify \( a \) and \( r \)**:
- \( a = x^3 \)
- \( r = \frac{x^5}{x^3} = x^2 \)

2. **Apply the formula**:
\[ S_n = \frac{x^3(1 - (x^2)^n)}{1 - x^2} \]

#### (ii) \( 1, -a, a^2, -a^3, \ldots \) (n terms)

1. **Identify \( a \) and \( r \)**:
- \( a = 1 \)
- \( r = -a \)

2. **Apply the formula**:
\[ S_n = \frac{1(1 - (-a)^n)}{1 - (-a)} = \frac{1 - (-a)^n}{1 + a} \]

#### (iii) \( x^2 - y^2, x - y, \frac{x - y}{x + y}, \ldots \) (n terms)

1. **Identify \( a \) and \( r \)**:
- \( a = x^2 - y^2 \)
- \( r = \frac{x - y}{x^2 - y^2} = \frac{1}{x + y} \)

2. **Apply the formula**:
\[ S_n = \frac{(x^2 - y^2)(1 - (\frac{1}{x + y})^n)}{1 - \frac{1}{x + y}} = \frac{(x^2 - y^2)(1 - (\frac{1}{x + y})^n)(x + y)}{(x + y - 1)} \]