Q.If the sum of $p$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$, show that the sum of $p + q$ terms is $-(p + q)$. Also, find the sum of first $p - q$ terms $(p > q)$.

Key Idea: Sum of $n$ terms of an Arithmetic Progression (AP).

Given $S_p = q$ and $S_q = p$:

1. $S_p = \frac{p}{2}[2a + (p-1)d] = q \implies 2ap + p(p-1)d = 2q$
2. $S_q = \frac{q}{2}[2a + (q-1)d] = p \implies 2aq + q(q-1)d = 2p$

Subtracting the second equation from the first:

$2a(p-q) + [p(p-1) - q(q-1)]d = 2(q-p)$

$2a(p-q) + [(p^2-q^2) - (p-q)]d = -2(p-q)$

$2a(p-q) + (p-q)(p+q-1)d = -2(p-q)$

Dividing by $(p-q)$ (since $p \neq q$):

$2a + (p+q-1)d = -2 \quad (*)$

The sum of $p+q$ terms is $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$.

Substituting $(*)$ into this expression:

$S_{p+q} = \frac{p+q}{2}(-2) = -(p+q)$.

To find the sum of $p-q$ terms, we first find $d$. Substitute $2a = -2 - (p+q-1)d$ from $(*)$ into $S_p = q$:

$\frac{p}{2}[-2 - (p+q-1)d + (p-1)d] = q$

$\frac{p}{2}[-2 + d(-p-q+1+p-1)] = q$

$\frac{p}{2}[-2 - qd] = q \implies -p - \frac{pqd}{2} = q \implies -\frac{pqd}{2} = p+q \implies d = -\frac{2(p+q)}{pq}$. …

Concept and Intuition

The core idea here is to leverage the given information about the sums of an Arithmetic Progression (A.P.) to determine its fundamental properties: the first term ($a$) and the common difference ($d$). Once we have expressions for $a$ and $d$ (or, more efficiently, for the term $2a + (n-1)d$), we can calculate the sum of any number of terms.

The problem provides two conditions:

1. The sum of $p$ terms is $q$.
2. The sum of $q$ terms is $p$.

We will translate these conditions into algebraic equations using the standard formula for the sum of $n$ terms of an A.P. This will give us a system of two linear equations in terms of $a$ and $d$. Solving this system will allow us to find $d$ and an expression involving $a$. Crucially, for calculating the sum of $N$ terms, $S_N = \frac{N}{2}[2a + (N-1)d]$, we don't always need to find $a$ and $d$ separately. Often, finding the expression $2a + (N-1)d$ directly is simpler and less prone to algebraic errors.

Step-by-step Solution

1. Recall the Sum Formula for an A.P.

   Let the first term of the A.P. be $a$ and the common difference be $d$. The sum of the first $n$ terms, denoted by $S_n$, is given by:

   $S_n = \frac{n}{2}[2a + (n-1)d]$

2. Formulate Equations from Given Conditions

   We are given two conditions:

   • The sum of $p$ terms is $q$:

    Multiplying by $2/p$ (assuming $p \neq 0$, which must be true for $p$ terms to exist):

*   The sum of $q$ terms is $p$:

    Multiplying by $2/q$ (assuming $q \neq 0$, which must be true for $q$ terms to exist):

3. Solve for the Common Difference ($d$) To find $d$, we can subtract Equation 2 from Equation 1. This eliminates the $2a$ term:

Since $p \neq q$ (otherwise $S_p=q$ and $S_q=p$ would imply $p=q$ and $S_p=p$, which is a trivial case, and the problem asks for $p-q$ terms with $p>q$), we can divide both sides by $(p-q)$:

4. Calculate the Sum of $p+q$ Terms ($S_{p+q}$) We need to find $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$. Let's focus on the term $2a + (p+q-1)d$. We can rewrite $(p+q-1)d$ as $(p-1)d + qd$. So, $2a + (p+q-1)d = [2a + (p-1)d] + qd$. From Equation 1, we know $2a + (p-1)d = \frac{2q}{p}$. Substitute this and the value of $d$: …