Discussion

Explanation:

Let the first term be a and common difference be d.

⇒ Tx = a + (x - 1)d, and
⇒ Ty = a + (y - 1)d

Given, xTx = yTy

⇒ x(a + (x - 1)d) = y(a + (y-1)d)
⇒ ax + x2d - xd = ay + y2d - yd
⇒ a(x - y)  + d(x2 - y2) - d(x - y) = 0
⇒ (x - y)(a + (x + y)d - d) = 0
⇒ a + (x + y - 1)d = 0 = Tx+y

∴ (x + y)th term is zero.

Hence, option (b).

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