Question

If the sum of the first p term of an ap is the same as the sum of its first q terms then show that the sum of its p+q term is zero​

Answer 1

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$let \: \: a \: \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: ap \: \\ then \\ s _{p} = s _{q} \implies \frac{p}{2} (2a + (p - 1)d) = \frac{q}{2} (2a + (q - 1)d \\ \implies(p - q)(2a) = (q - p)(q + p - 1) \\ \implies2a = (1 - p - q)d \: \: \: \: \: \: \: .....(1) \\ sum \: of \: the \: first \: (p + q) \: terms \: of \: the \: given \: ap \\ = \frac{(p + q)}{2} (2a + (p + q - 1)d) \\ = \frac{(p + q)}{2} .(1 - p - q)d + (p + q - 1)d \: \: \: \: \: \: \: \: (using \: 1) \\ = 0$

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Answer 2

Answer:

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Step-by-step explanation:

Sum of first p terms of an AP is�

Sum of first q terms of an AP is�

Sum of n terms of an AP =n2(2a+(n−1)d)

So,  

Sum of p terms of an AP =p2(2a+(p−1)d) and

Sum of q terms of an AP =q2(2a+(q−1)d) are equal. (Given)

Therefore, 2ap+p(p−1)d=2aq+q(q−1)d

2a(p−q)=d(q(q−1)−p(p−1))

2a(p−q)=d(q2−q−p2+p)

2a(p−q)=d((q2−p2)+(p−q))

2a(p−q)=d((q−p)(q+p)+(p−q))

2a(p−q)=d(p−q)(1−p−q)

2a=−d(p+q−1)(Equation 1)

Sum of (p+q) terms of an AP =p+q2(2a+(p+q−1)d)

Substituting value of 2a from Equation 1, we get,

Sum  =p+q2(−d(p+q−1)+d(p+q−1))=0