Question

BEST GETS BRAINLIEST :
If the sum of first p terms of an A.P. = the sum of first q terms then show that the sum of its first (p+q) terms is zero. (p ≠ q )

Answer 1

The answer is given below :

Let us consider that the first term of the AP is a and the common difference for the AP is d.

Then, the sum of the first p terms

= p/2 × [2a + (p - 1)d]

and the sum of the first q terms

= q/2 × [2a + (q - 1)d]

By the given condition,

the sum of the first p terms = the sum of the first q terms

⇒ p/2 × [2a + (p - 1)d] = q/2 × [2a + (q - 1)d]

⇒ p × (2a + pd - d) = q × (2a + qd - d)

⇒ 2ap + p²d - pd = 2aq + q²d - qd

⇒ 2a(p - q) + (p² - q²)d - (p - q)d = 0

⇒ 2a(p - q) + (p + q)(p - q)d - (p - q)d = 0,
using the identity a² - b² = (a + b)(a - b)

⇒ 2a + (p + q)d - d = 0, since (p - q) ≠ 0

⇒ 2a + (p + q - 1)d = 0 .....(i)

Now, the sum of the first (p + q) term

= (p + q)/2 × [2a + (p + q - 1)d]

= (p + q)/2 × 0, by (i)

= 0 [Proved]

RULE :

We know that, pth term
= first term + (p - 1)×common difference

= a + (p - 1)d

Sum of p terms

= p/2 × (first term + pth term)

= p/2 × [a + {a + (p - 1)d}]

= p/2 × [a + a + (p - 1)d]

= p/2 × {2a + (p - 1)d}

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