Discussion

Explanation:

Let the first term of the series = a and common difference = d.

Given that, T_p = a + (p – 1)d = q            …(1)

And T_q = a + (q – 1)d = p            …(2)

(2) - (1), we get

(p - q)d = q - p

⇒ d = $\frac{q-p}{(p-q) }$ = -1

Putting the value of d in equation (1), then a = p + q – 1 

Now r^th term is given by A.P.

T_r = a + (r - 1)d = (p + q - 1) + (r - 1)(-1) = p + q – r.

Alternately,
Put p = 1, q = 2 and r = 3.

∴ T₁ = 2 and T2 = 1.

⇒ a = 2 and d = -1.

∴ T₃ = 2 + (3 - 1) × -1 = 0

Use options now. By substituting p = 1, q = 2 and r = 3 and see which options gives us T₃ = 0.

Option (b) gives us p + q – r = 1 + 2 – 3 = 0.

Hence, option (b).

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