The sums of n terms of three arithmetical progressions are `S_(1), S_(2) and S_(3)`. The first term of each is unity and the common differences are 1, 2 and 3 respectively. Then `S_(1) + S_(3) = 3S_(2)`

Answer

To solve the problem, we need to find the sums of the first n terms of three arithmetic progressions (APs) and verify the equation \( S_1 + S_3 = 3S_2 \).

### Step 1: Define the sums of the arithmetic progressions

1. **For the first AP**:
- First term \( a_1 = 1 \)
- Common difference \( d_1 = 1 \)
- The sum of the first n terms \( S_1 \) is given by the formula:
\[ S_1 = \frac{n}{2} \times [2a_1 + (n-1)d_1] = \frac{n}{2} \times [2 \cdot 1 + (n-1) \cdot 1] \]
- Simplifying this:
\[ S_1 = \frac{n}{2} \times [2 + n - 1] = \frac{n}{2} \times (n + 1) = \frac{n(n + 1)}{2} \]

2. **For the second AP**:
- First term \( a_2 = 1 \)
- Common difference \( d_2 = 2 \)
- The sum of the first n terms \( S_2 \) is given by:
\[ S_2 = \frac{n}{2} \times [2a_2 + (n-1)d_2] = \frac{n}{2} \times [2 \cdot 1 + (n-1) \cdot 2] \]
- Simplifying this:
\[ S_2 = \frac{n}{2} \times [2 + 2n - 2] = \frac{n}{2} \times 2n = n^2 \]

3. **For the third AP**:
- First term \( a_3 = 1 \)
- Common difference \( d_3 = 3 \)
- The sum of the first n terms \( S_3 \) is given by:
\[ S_3 = \frac{n}{2} \times [2a_3 + (n-1)d_3] = \frac{n}{2} \times [2 \cdot 1 + (n-1) \cdot 3] \]
- Simplifying this:
\[ S_3 = \frac{n}{2} \times [2 + 3n - 3] = \frac{n}{2} \times (3n - 1) = \frac{n(3n - 1)}{2} \]

### Step 2: Verify the equation \( S_1 + S_3 = 3S_2 \)

Now we need to check if \( S_1 + S_3 = 3S_2 \):

1. **Calculate \( S_1 + S_3 \)**:
\[ S_1 + S_3 = \frac{n(n + 1)}{2} + \frac{n(3n - 1)}{2} \]
- Combine the fractions:
\[ S_1 + S_3 = \frac{n(n + 1) + n(3n - 1)}{2} = \frac{n(n + 1 + 3n - 1)}{2} = \frac{n(4n)}{2} = 2n^2 \]

2. **Calculate \( 3S_2 \)**:
\[ 3S_2 = 3(n^2) = 3n^2 \]

### Step 3: Compare \( S_1 + S_3 \) and \( 3S_2 \)

We have:
- \( S_1 + S_3 = 2n^2 \)
- \( 3S_2 = 3n^2 \)

Clearly, \( S_1 + S_3 \neq 3S_2 \).

### Conclusion

Thus, the statement \( S_1 + S_3 = 3S_2 \) is **false**.