If 1,2,3….are first terms, 1,3,5….are common differences and `S_(1),S_(2),S_(3)`…….are sums of n terms of given p AP's, then `S_(1)+S_(2)+S_(3)+…….+S_(p)` is equal to

Options

A. `(np(np+1))/(2)`
B. `(n(np+1))/(2)`
C. `(np(p+1))/(2)`
D. `(np(np-1))/(2)`.

Answer

To solve the problem, we need to find the sum \( S_1 + S_2 + S_3 + \ldots + S_P \) where \( S_n \) is the sum of the first \( n \) terms of an arithmetic progression (AP) defined by the first term and common difference.

### Step-by-step Solution:

1. **Identify the First Terms and Common Differences**:
- The first terms of the APs are given as \( 1, 2, 3, \ldots, P \).
- The common differences are given as \( 1, 3, 5, \ldots \). The \( n \)-th common difference can be expressed as \( 2n - 1 \).

2. **Formula for the Sum of n Terms of an AP**:
The sum of the first \( n \) terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms.

3. **Calculate \( S_1, S_2, S_3, \ldots, S_P \)**:
- For \( S_1 \):
\[ S_1 = \frac{n}{2} \left(2 \cdot 1 + (n-1) \cdot 1\right) = \frac{n}{2} \left(2 + n - 1\right) = \frac{n}{2} (n + 1) \]
- For \( S_2 \):
\[ S_2 = \frac{n}{2} \left(2 \cdot 2 + (n-1) \cdot 3\right) = \frac{n}{2} \left(4 + 3(n - 1)\right) = \frac{n}{2} (3n + 1) \]
- For \( S_3 \):
\[ S_3 = \frac{n}{2} \left(2 \cdot 3 + (n-1) \cdot 5\right) = \frac{n}{2} \left(6 + 5(n - 1)\right) = \frac{n}{2} (5n + 1) \]
- Continuing this way, we can generalize:
\[ S_k = \frac{n}{2} \left(2k + (n-1)(2k - 1)\right) \]

4. **Sum All \( S_k \)**:
We need to find:
\[ S_1 + S_2 + S_3 + \ldots + S_P \]
This can be expressed as:
\[ \sum_{k=1}^{P} S_k = \sum_{k=1}^{P} \frac{n}{2} \left(2k + (n-1)(2k - 1)\right) \]
Simplifying this sum gives us:
\[ = \frac{n}{2} \sum_{k=1}^{P} \left(2k + (n-1)(2k - 1)\right) \]

5. **Calculate the Summation**:
The summation can be split into two parts:
\[ = \frac{n}{2} \left(2\sum_{k=1}^{P} k + (n-1)(2\sum_{k=1}^{P} k - P)\right) \]
Using the formula for the sum of the first \( P \) natural numbers \( \sum_{k=1}^{P} k = \frac{P(P+1)}{2} \), we can substitute and simplify.

6. **Final Result**:
After simplifying, we will find:
\[ S_1 + S_2 + S_3 + \ldots + S_P = \frac{nP(P+1)}{4} \]

### Final Answer:
\[ S_1 + S_2 + S_3 + \ldots + S_P = \frac{nP(P+1)}{4} \]