Question

if the sum of first p terms of an AP is the same as the
Sum of its first q terms where p is not equal to q, then show that the sum of its first (p +q) terms is zero.
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Answer 1

Answer:

Step-by-step explanation:

let the first term of an AP is 'a' and common difference is 'd'.

sum of p terms = (p/2)(2a + (p-1)d)

sum of q terms = (q/2)(2a + (q-1)d)

sum of (p+q) terms = ((p+q)/2)(2a + (p+q-1)d)

now since sum of p terms = sum of q terms

therefore,

(p/2)(2a + (p-1)d) = (q/2)(2a + (q-1)d)

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

2pa + p²d - pd = 2aq + q²d - qd

d(q-p) = 2aq + q²d - 2pa - p²d

        = 2a(q-p) +d(q²-p²)

        = (q-p)(2a +d(q+p))

d = 2a + dq + dp

d(1-p-q) = 2a

d = 2a/(1-p-q)

putting this value of d in the sum of p+q terms, we get,

((p+q)/2)(2a + (p+q-1)d)

((p+q)/2)(2a + (p+q-1)(2a/(1-p-q))) = ((p+q)/2)(2a - (-p-q+1)(2a/(1-p-q)))

((p+q)/2)(2a - 2a)

=0

Answer 2

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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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