Understanding the Problem: Arithmetic Progression (A.P.)

The problem provides information about the sum of terms in an Arithmetic Progression (A.P.). We are given the sum of the first \(n\) terms, denoted as \(S_n\), and the sum of the first \(m\) terms, denoted as \(S_m\). Specifically, we have:

• \(S_n = pn^2\)
• \(S_m = pm^2\)

Our goal is to find the sum of the first \(p\) terms, \(S_p\).

Key Formula for Sum of A.P. Terms

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

\(S_r = \dfrac{r}{2} [2a + (r-1)d]\)

where:

• \(S_r\) is the sum of the first \(r\) terms
• \(a\) is the first term of the A.P.
• \(d\) is the common difference of the A.P.

Setting up Equations from Given Information

Using the formula for \(S_r\) and the given information, we can set up two equations:

For \(S_n\):

\(\dfrac{n}{2} [2a + (n-1)d] = pn^2\)

Multiplying both sides by \(\dfrac{2}{n}\) (assuming \(n \neq 0\)), we get:

\(2a + (n-1)d = 2pn \quad \cdots (1)\)

For \(S_m\):

\(\dfrac{m}{2} [2a + (m-1)d] = pm^2\)

Multiplying both sides by \(\dfrac{2}{m}\) (assuming \(m \neq 0\)), we get:

\(2a + (m-1)d = 2pm \quad \cdots (2)\)

Solving for the First Term (\(a\)) and Common Difference (\(d\))

We now have a system of two linear equations with two variables, \(a\) and \(d\). We can solve this system to find the values of \(a\) and \(d\).

Subtract equation (2) from equation (1):

\((2a + (n-1)d) - (2a + (m-1)d) = 2pn - 2pm\)

Simplify the left side:

\(2a + nd - d - 2a - md + d = 2p(n-m)\)

\(nd - md = 2p(n-m)\)

\((n-m)d = 2p(n-m)\)

Assuming \(n \neq m\), we can divide both sides by \((n-m)\):

\(d = 2p\)

Now substitute the value of \(d = 2p\) into equation (1):

\(2a + (n-1)(2p) = 2pn\)

\(2a + 2pn - 2p = 2pn\)

Subtract \(2pn\) from both sides:

\(2a - 2p = 0\)

\(2a = 2p\)

\(a = p\)

So, the first term of the A.P. is \(p\) and the common difference is \(2p\).

Calculating the Sum of the First \(p\) Terms (\(S_p\))

Now that we have the first term \(a=p\) and the common difference \(d=2p\), we can find the sum of the first \(p\) terms using the formula for \(S_r\) with \(r=p\).

\(S_p = \dfrac{p}{2} [2a + (p-1)d]\)

Substitute \(a=p\) and \(d=2p\):

\(S_p = \dfrac{p}{2} [2(p) + (p-1)(2p)]\)

\(S_p = \dfrac{p}{2} [2p + 2p^2 - 2p]\)

Simplify inside the brackets:

\(S_p = \dfrac{p}{2} [2p^2]\)

Multiply:

\(S_p = p \cdot p^2\)

\(S_p = p^3\)

Conclusion

The sum of the first \(p\) terms of the A.P. is \(p^3\).

Revision Table: Important A.P. Formulas

Here's a quick look at the key formulas used when working with Arithmetic Progressions:

Additional Information: Arithmetic Progressions Explained

An Arithmetic Progression (A.P.) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by \(d\).

Examples of A.P.s include:

• 2, 4, 6, 8, ... (common difference \(d=2\))
• 10, 7, 4, 1, ... (common difference \(d=-3\))
• 5, 5, 5, 5, ... (common difference \(d=0\))

The first term of an A.P. is usually denoted by \(a\) or \(a_1\).

The general term, or the \(n\)-th term, of an A.P. can be found using the formula \(a_n = a + (n-1)d\). This formula helps us find any term in the sequence if we know the first term and the common difference.

The sum of the first \(n\) terms, \(S_n\), is useful for finding the total value when adding up a certain number of terms in the progression. The formula \(S_n = \dfrac{n}{2} (2a + (n-1)d)\) is derived by averaging the first and last term and multiplying by the number of terms, or by summing the series \(a + (a+d) + \dots + (a+(n-1)d)\) directly.

In this problem, we used the given sum formulas \(S_n\) and \(S_m\) to work backward and find the fundamental properties (first term and common difference) of the A.P. Once these properties were known, we could calculate the sum of any number of terms, specifically \(S_p\). The structure of the given sum formulas \(S_r = pr^2\) is specific and implies a particular type of A.P. where \(a=p\) and \(d=2p\).