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Sum of n terms of AP is given as:

Sum of p terms, S_p is given as:

Sum of q terms, S_q is given as:

Now, its given that S_p = S_q

⇒

⇒ 2ap + pd(p - 1) = 2aq + qd(q - 1)

⇒ 2a(p - q) + d(p² - q²) = (p - q)d

⇒ 2a(p - q) + d(p - q)(p + q) = (p - q)d

⇒ 2a + d(p + q) = d

⇒ 2a = - d(p + q - 1) ……………..(1)

Sum of (p + q) terms :

S_{p + q} =

From eq(1) putting value of 2a,

S_{p + q} =

S_{p + q} = 0.

Hence proved.