If `S_(k)` denotes the sum of first k terms of a G.P. Then, `S_(n),S_(2n)-S_(n),S_(3n)-S_(2n)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Explanation

To solve the problem, we need to analyze the sums of the first \( n \), \( 2n \), and \( 3n \) terms of a geometric progression (G.P.). Let's denote the first term of the G.P. as \( a \) and the common ratio as \( r \).

### Step 1: Write the formula for the sum of the first \( k \) terms of a G.P.
The formula for the sum of the first \( k \) terms of a G.P. is given by:

\[ S_k = \frac{a(r^k - 1)}{r - 1} \]

### Step 2: Calculate \( S_n \), \( S_{2n} \), and \( S_{3n} \)
Using the formula, we can find:

1. **Sum of the first \( n \) terms**:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]

2. **Sum of the first \( 2n \) terms**:
\[ S_{2n} = \frac{a(r^{2n} - 1)}{r - 1} \]

3. **Sum of the first \( 3n \) terms**:
\[ S_{3n} = \frac{a(r^{3n} - 1)}{r - 1} \]

### Step 3: Calculate \( S_{2n} - S_n \)
Now, we need to find \( S_{2n} - S_n \):

\[ S_{2n} - S_n = \frac{a(r^{2n} - 1)}{r - 1} - \frac{a(r^n - 1)}{r - 1} \]

Combining the fractions gives:

\[ S_{2n} - S_n = \frac{a(r^{2n} - 1 - (r^n - 1))}{r - 1} = \frac{a(r^{2n} - r^n)}{r - 1} \]

Factoring out \( r^n \):

\[ S_{2n} - S_n = \frac{a r^n (r^n - 1)}{r - 1} \]

### Step 4: Calculate \( S_{3n} - S_{2n} \)
Next, we calculate \( S_{3n} - S_{2n} \):

\[ S_{3n} - S_{2n} = \frac{a(r^{3n} - 1)}{r - 1} - \frac{a(r^{2n} - 1)}{r - 1} \]

Combining the fractions gives:

\[ S_{3n} - S_{2n} = \frac{a(r^{3n} - r^{2n})}{r - 1} \]

Factoring out \( r^{2n} \):

\[ S_{3n} - S_{2n} = \frac{a r^{2n} (r^n - 1)}{r - 1} \]

### Step 5: Determine the relationship between \( S_n \), \( S_{2n} - S_n \), and \( S_{3n} - S_{2n} \)
Now we have:

1. \( S_n = \frac{a(r^n - 1)}{r - 1} \)
2. \( S_{2n} - S_n = \frac{a r^n (r^n - 1)}{r - 1} \)
3. \( S_{3n} - S_{2n} = \frac{a r^{2n} (r^n - 1)}{r - 1} \)

### Step 6: Check if these three terms are in G.P.
To check if \( S_n \), \( S_{2n} - S_n \), and \( S_{3n} - S_{2n} \) are in G.P., we need to verify if:

\[ \frac{S_{2n} - S_n}{S_n} = \frac{S_{3n} - S_{2n}}{S_{2n} - S_n} \]

Calculating the left-hand side:

\[ \frac{S_{2n} - S_n}{S_n} = \frac{\frac{a r^n (r^n - 1)}{r - 1}}{\frac{a(r^n - 1)}{r - 1}} = r^n \]

Calculating the right-hand side:

\[ \frac{S_{3n} - S_{2n}}{S_{2n} - S_n} = \frac{\frac{a r^{2n} (r^n - 1)}{r - 1}}{\frac{a r^n (r^n - 1)}{r - 1}} = r^n \]

Since both sides are equal, we conclude that \( S_n \), \( S_{2n} - S_n \), and \( S_{3n} - S_{2n} \) are in G.P.

### Final Answer
Thus, the answer is that \( S_n, S_{2n} - S_n, S_{3n} - S_{2n} \) are in G.P.

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