The problem asks us to find a root of a given quadratic equation, given properties of an Arithmetic Progression (A.P.). Let the A.P. have the first term '$a$' and common difference '$d$'. The $n$th term is given by $t_n = a + (n-1)d$.

Finding A.P. Parameters

We are given:

• $t_p = \frac{1}{q} \implies a + (p-1)d = \frac{1}{q}$ (Equation 1)
• $t_q = \frac{1}{p} \implies a + (q-1)d = \frac{1}{p}$ (Equation 2)

Subtracting Equation 2 from Equation 1:

$(a + (p-1)d) - (a + (q-1)d) = \frac{1}{q} - \frac{1}{p}$ $(p-1 - q + 1)d = \frac{p-q}{pq}$ $(p-q)d = \frac{p-q}{pq}$

Assuming $p \neq q$, we divide by $(p-q)$ to get the common difference:

$d = \frac{1}{pq}$

Substitute $d = \frac{1}{pq}$ into Equation 1:

$a + (p-1)\frac{1}{pq} = \frac{1}{q}$ $a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q}$ $a + \frac{1}{q} - \frac{1}{pq} = \frac{1}{q}$ $a = \frac{1}{pq}$

So, the first term $a = \frac{1}{pq}$ and the common difference $d = \frac{1}{pq}$.

Evaluating Option A: $t_{pq}$

Let's find the value of the term $t_{pq}$:

$t_{pq} = a + (pq-1)d$ $t_{pq} = \frac{1}{pq} + (pq-1)\frac{1}{pq}$ $t_{pq} = \frac{1}{pq} + 1 - \frac{1}{pq}$ $t_{pq} = 1$

Testing Roots of the Quadratic Equation

The given quadratic equation is:

$(p+2q-3r)x^2 + (q+2r-3p)x + (r+2p-3q) = 0$

Let's test if $x=1$ is a root. Substitute $x=1$ into the equation:

$(p+2q-3r)(1)^2 + (q+2r-3p)(1) + (r+2p-3q)$ $= (p+2q-3r) + (q+2r-3p) + (r+2p-3q)$

Combine the coefficients for $p$, $q$, and $r$:

$p: (1 - 3 + 2) = 0$ $q: (2 + 1 - 3) = 0$ $r: (-3 + 2 + 1) = 0$

The sum is $0p + 0q + 0r = 0$. Therefore, $x=1$ is a root of the quadratic equation.

Conclusion

Since we found $t_{pq} = 1$ and $x=1$ is a root of the equation, $t_{pq}$ is a root of the equation.