Question:

Step 1: Set up the arithmetic progression conditions. For an arithmetic progression with first term $a$ and common difference $d$, we have: $$T_p = a + (p-1)d = \frac{1}{q}$$ $$T_q = a + (q-1)d = \frac{1}{p}$$ Step 2: Find the common difference $d$. Subtract the first equation from the second to eliminate $a$: $$(q-1)d - (p-1)d = \frac{1}{p} - \frac{1}{q}$$ $$(q-p)d = \frac{q-p}{pq}$$ Since $p \neq q$, we can divide both sides by $(q-p)$: $$d = \frac{1}{pq}$$ Step 3: Find the first term $a$. Substitute $d = \frac{1}{pq}$ into the first equation: $$a + (p-1) \cdot \frac{1}{pq} = \frac{1}{q}$$ $$a = \frac{1}{q} - \frac{p-1}{pq} = \frac{p - (p-1)}{pq} = \frac{1}{pq}$$ Step 4: Find the general term $T_n$ and calculate $T_{pq}$. The general term of the AP is: $$T_n = a + (n-1)d = \frac{1}{pq} + (n-1) \cdot \frac{1}{pq} = \frac{1 + (n-1)}{pq} = \frac{n}{pq}$$ Therefore: $$T_{pq} = \frac{pq}{pq} = 1$$ Step 5: Verify that $x = 1$ is a root of the given equation. Substitute $x = 1$ into the equation $(p+2q-3r)x^2 + (q+2r-3p)x + (r+2p-3q) = 0$: $$(p+2q-3r)(1)^2 + (q+2r-3p)(1) + (r+2p-3q)$$ $$= p + 2q - 3r + q + 2r - 3p + r + 2p - 3q$$ $$= (p - 3p + 2p) + (2q + q - 3q) + (-3r + 2r + r)$$ $$= 0 + 0 + 0 = 0$$ So $x = 1$ is always a root of the equation, regardless of the values of $p$, $q$, and $r$. Step 6: Conclude that $T_{pq}$ is necessarily a root. Since $T_{pq} = 1$ and $x = 1$ is necessarily a root of the given equation, $T_{pq}$ is necessarily a root. **Final Answer: Option 1** ($T_{pq}$)