Step 1: Understanding the Concept:
This question has two unrelated-looking components. One is a property of an Arithmetic Progression (A.P.) and the other is about the roots of a quadratic equation. The key is to notice that if the sum of coefficients of a quadratic equation \( ax^2 + bx + c = 0 \) is zero, then \( x = 1 \) is always a root. We then need to check which A.P. term equals 1.
Step 2: Key Formula or Approach:
1. A.P. Term: \( t_n = a + (n - 1)d \).
2. Quadratic Root: If \( a + b + c = 0 \), then \( x = 1 \) is a root.
Step 3: Detailed Explanation:
Let the first term of the A.P. be \( a \) and the common difference be \( d \).
Given:
\( t_p = a + (p - 1)d = \frac{1}{q} \) \dots (1)
\( t_q = a + (q - 1)d = \frac{1}{p} \) \dots (2)
Subtracting (2) from (1):
\( (p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq} \implies d = \frac{1}{pq} \).
Substitute \( d \) into (1):
\( a + (p - 1)\frac{1}{pq} = \frac{1}{q} \implies a = \frac{1}{q} - \frac{p - 1}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq} \).
So, \( a = \frac{1}{pq} \) and \( d = \frac{1}{pq} \).
Now, calculate \( t_{pq} \):
\( t_{pq} = a + (pq - 1)d = \frac{1}{pq} + (pq - 1)\frac{1}{pq} = \frac{1 + pq - 1}{pq} = \frac{pq}{pq} = 1 \).
Now consider the quadratic equation:
\( (p + 2q - 3r)x^2 + (q + 2r - 3p)x + (r + 2p - 3q) = 0 \).
Sum the coefficients:
Sum \( = (p + 2q - 3r) + (q + 2r - 3p) + (r + 2p - 3q) \)
Sum \( = (p - 3p + 2p) + (2q + q - 3q) + (-3r + 2r + r) = 0 + 0 + 0 = 0 \).
Since the sum of coefficients is 0, \( x = 1 \) is a root.
Since \( t_{pq} = 1 \), it follows that \( t_{pq} \) is a root.
Step 4: Final Answer:
The root of the equation is 1, and the \( pq \)-th term of the given progression is also 1. Thus, \( t_{pq} \) is the correct option.