Understanding the Arithmetic Progression Problem

This question involves an Arithmetic Progression (AP), a sequence of numbers where the difference between consecutive terms is constant. We are given a specific condition relating the pth term and the qth term of an AP and asked to find the value of the (p+q)th term.

Key Concepts for AP

• The formula for the nth term of an AP is:
  \(a_n = a + (n-1)d\) where 'a' is the first term and 'd' is the common difference.

Analyzing the Given Condition

The problem states that 'p times the pth term is equal to q times the qth term'. We can write this mathematically as:

\( p \times a_p = q \times a_q \quad (\text{where } p \neq q) \)

Step-by-Step Algebraic Solution

1. Express terms using the AP formula:
   The pth term (\(a_p\)) is \(a + (p-1)d\).
   The qth term (\(a_q\)) is \(a + (q-1)d\).
2. Substitute into the given condition:
   \( p[a + (p-1)d] = q[a + (q-1)d] \)
3. Expand both sides of the equation:
   \( pa + p(p-1)d = qa + q(q-1)d \) \( pa + (p^2 - p)d = qa + (q^2 - q)d \)
4. Rearrange the terms to group 'a' and 'd':
   \( pa - qa = (q^2 - q)d - (p^2 - p)d \) \( a(p - q) = [q^2 - q - p^2 + p]d \)
5. Factorize the terms involving \(p^2\) and \(q^2\):
   \( a(p - q) = [(q^2 - p^2) - (q - p)]d \) \( a(p - q) = [-(p^2 - q^2) + (p - q)]d \) Using the difference of squares formula (\(p^2 - q^2 = (p-q)(p+q)\)): \( a(p - q) = [-(p-q)(p+q) + (p - q)]d \)
6. Simplify by dividing by \((p-q)\):
   Since we are given that \(p \neq q\), the term \((p-q)\) is not zero. We can divide both sides by \((p-q)\): \( a = [-(p+q) + 1]d \) \( a = (1 - p - q)d \)
7. Calculate the \((p+q)\)th term:
   We need to find \(a_{p+q}\). Using the AP formula \(a_n = a + (n-1)d\): \( a_{p+q} = a + ((p+q)-1)d \) Now, substitute the expression we found for 'a': \( a_{p+q} = [(1 - p - q)d] + (p+q-1)d \) \( a_{p+q} = (1 - p - q + p + q - 1)d \) \( a_{p+q} = (0)d \) \( a_{p+q} = 0 \)

Conclusion

Based on the algebraic derivation, if p times the pth term of an AP is equal to q times the qth term (where \(p \neq q\)), then the (p+q)th term of the AP is always 0.