To solve the problem step by step, we will find the sum of the terms of all the given arithmetic progressions (APs).
### Step 1: Identify the first terms and common differences of the APs
The first terms of the p APs are:
- 1, 2, 3, ..., p
The common differences of the p APs are:
- 1, 3, 5, ..., (2p - 1)
### Step 2: Calculate the sum of the first terms
The sum of the first terms can be calculated using the formula for the sum of the first n natural numbers:
\[
S_A = \sum_{k=1}^{p} k = \frac{p(p + 1)}{2}
\]
### Step 3: Calculate the sum of the common differences
The common differences form an arithmetic progression where:
- First term (A) = 1
- Common difference (d) = 2
- Number of terms (n) = p
The last term of this AP is \(2p - 1\). The sum of the first p odd numbers can also be calculated as:
\[
S_D = \sum_{k=1}^{p} (2k - 1) = p^2
\]
### Step 4: Calculate the sum of the terms of each AP
The sum of the terms of an AP is given by the formula:
\[
S_n = \frac{n}{2} \left(2A + (n - 1)d\right)
\]
For each AP, we have:
- \(A\) as the first term (1, 2, 3, ..., p)
- \(d\) as the common difference (1, 3, 5, ..., (2p - 1))
- \(n\) as the number of terms (n)
Thus, for each AP, the sum becomes:
\[
S_{AP} = \frac{n}{2} \left(2k + (n - 1)(2k - 1)\right)
\]
where \(k\) is the first term of the respective AP.
### Step 5: Substitute the values into the sum formula
Substituting the values of \(A\) and \(d\):
\[
S_{AP} = \frac{n}{2} \left(2k + (n - 1)(2k - 1)\right)
\]
This simplifies to:
\[
S_{AP} = \frac{n}{2} \left(2k + (n - 1)(2k) - (n - 1)\right)
\]
\[
= \frac{n}{2} \left(n \cdot 2k - (n - 1) + 2k\right)
\]
\[
= \frac{n}{2} \left( (n + 1) \cdot 2k - (n - 1) \right)
\]
### Step 6: Total sum of all APs
Now, we need to sum \(S_{AP}\) for all \(k\) from 1 to \(p\):
\[
S_{total} = \sum_{k=1}^{p} S_{AP}
\]
\[
= \sum_{k=1}^{p} \frac{n}{2} \left( (n + 1) \cdot 2k - (n - 1) \right)
\]
This can be simplified to:
\[
= \frac{n}{2} \left( (n + 1) \cdot \sum_{k=1}^{p} 2k - p(n - 1) \right)
\]
Using the sum of the first \(p\) natural numbers:
\[
= \frac{n}{2} \left( (n + 1) \cdot p(p + 1) - p(n - 1) \right)
\]
### Final Result
Thus, the total sum of the terms of all the progressions is:
\[
= \frac{np}{2} \left( nP + 1 \right)
\]
