Given p A.P's , each of which consists on n terms . If their first terms are `1,2,3……` p and common differences are `1,3,5,……,2p-1 ` respectively , then sum of the terms of all the progressions is

Options

A. `1/ 2 np(np+1)`
B. `1/2 n(p+1)`
C. `np(n+1)`
D. none of these

Answer

To solve the problem step by step, we need to find the sum of the terms of \( p \) arithmetic progressions (A.P.s), each consisting of \( n \) terms, where the first terms and common differences of the A.P.s are given.

### Step 1: Identify the first terms and common differences
The first terms of the \( p \) A.P.s are:
- \( a_1 = 1 \)
- \( a_2 = 2 \)
- \( a_3 = 3 \)
- ...
- \( a_p = p \)

The common differences of the \( p \) A.P.s are:
- \( d_1 = 1 \)
- \( d_2 = 3 \)
- \( d_3 = 5 \)
- ...
- \( d_p = 2p - 1 \)

### Step 2: Use the formula for the sum of an A.P.
The sum \( S_n \) of the first \( n \) terms of an A.P. can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 3: Calculate the sum for each A.P.
For each \( k \)-th A.P. (where \( k = 1, 2, \ldots, p \)):
- First term \( a_k = k \)
- Common difference \( d_k = 2k - 1 \)

Thus, the sum of the \( k \)-th A.P. is:
\[ S_k = \frac{n}{2} \times (2k + (n - 1)(2k - 1)) \]
Expanding this:
\[ S_k = \frac{n}{2} \times (2k + (2nk - 2k - n + 1)) = \frac{n}{2} \times (2nk - n + 1) \]

### Step 4: Calculate the total sum of all A.P.s
Now, we need to sum \( S_k \) from \( k = 1 \) to \( p \):
\[ S = \sum_{k=1}^{p} S_k = \sum_{k=1}^{p} \frac{n}{2} \times (2nk - n + 1) \]

This can be simplified as:
\[ S = \frac{n}{2} \sum_{k=1}^{p} (2nk - n + 1) = \frac{n}{2} \left( 2n \sum_{k=1}^{p} k - pn + p \right) \]

### Step 5: Use the formula for the sum of the first \( p \) natural numbers
The sum \( \sum_{k=1}^{p} k \) is given by:
\[ \sum_{k=1}^{p} k = \frac{p(p + 1)}{2} \]

Substituting this into our equation gives:
\[ S = \frac{n}{2} \left( 2n \cdot \frac{p(p + 1)}{2} - pn + p \right) \]
\[ = \frac{n}{2} \left( np(p + 1) - pn + p \right) \]
\[ = \frac{n}{2} \left( np^2 + np - pn + p \right) \]
\[ = \frac{n}{2} \left( np^2 + p \right) \]

### Final Step: Simplify the expression
Thus, the total sum of the terms of all the A.P.s is:
\[ S = \frac{np^2 + np}{2} = \frac{np(p + 1)}{2} \]

### Conclusion
The final result for the sum of the terms of all the progressions is:
\[ S = \frac{np(p + 1)}{2} \]