If `S_(1), S_(2), S_(3)` be the sum of n, 2n, 3n terms respectively of an A.P., then

Options

A. `S_(3) = S_(1) + S_(2)`
B. `S_(3) = 2 (S_(1) + S_(2))`
C. `S_(3) = 3 (S_(2) - S_(1))`
D. none of these

Answer

To solve the problem, we need to find the relationship between the sums \( S_1, S_2, \) and \( S_3 \) of the first \( n, 2n, \) and \( 3n \) terms of an arithmetic progression (A.P.).

### Step 1: Write the formulas for \( S_1, S_2, \) and \( S_3 \)

The sum of the first \( n \) terms of an A.P. is given by the formula:

\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

Using this formula, we can write:

- For \( S_1 \) (sum of first \( n \) terms):
\[ S_1 = \frac{n}{2} \left(2a + (n-1)d\right) \]

- For \( S_2 \) (sum of first \( 2n \) terms):
\[ S_2 = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]

- For \( S_3 \) (sum of first \( 3n \) terms):
\[ S_3 = \frac{3n}{2} \left(2a + (3n-1)d\right) \]

### Step 2: Simplify the expressions

Now we will simplify \( S_1, S_2, \) and \( S_3 \):

1. **For \( S_1 \)**:
\[ S_1 = \frac{n}{2} (2a + (n-1)d) = \frac{n}{2} (2a + nd - d) = \frac{n}{2} (2a + nd - d) \]

2. **For \( S_2 \)**:
\[ S_2 = n (2a + (2n-1)d) = n (2a + 2nd - d) = 2an + n(2nd - d) \]

3. **For \( S_3 \)**:
\[ S_3 = \frac{3n}{2} (2a + (3n-1)d) = \frac{3n}{2} (2a + 3nd - d) = \frac{3n}{2} (2a + 3nd - d) \]

### Step 3: Find the relationship between \( S_1, S_2, \) and \( S_3 \)

Now, we can find the difference between \( S_2 \) and \( S_1 \):

\[ S_2 - S_1 = n(2a + 2nd - d) - \frac{n}{2}(2a + nd - d) \]

Factoring out \( n \):

\[ = n \left( (2a + 2nd - d) - \frac{1}{2}(2a + nd - d) \right) \]

### Step 4: Simplify the expression

Now, simplifying the expression inside the parentheses:

\[ = n \left( (2a + 2nd - d) - \left(a + \frac{nd}{2} - \frac{d}{2}\right) \right) \]

This will lead to:

\[ = n \left( \frac{3a}{2} + \frac{3nd}{2} \right) \]

### Step 5: Conclude the relationship

From the calculations, we can conclude that:

\[ S_3 = 3S_1 \]

Thus, the relationship among \( S_1, S_2, \) and \( S_3 \) is established.

### Final Answer
The relationship is \( S_3 = 3S_1 \).

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