Question

if the sum of P term of an ap is Q and the sum of Q terms is P show that sum of (p+q) terms is -(p+q)​

Answer 1

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Answer 2

Step-by-step explanation:

$let \: a \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: a.p. \\ then \: sp \: = \: sq \: \: \: \: \: \: ....(given) \\ using \: formula \: \\ sn = \frac{n}{2} (2a \: + (n - 1)d)\: \\ \frac{p}{2} (2a + (p - 1)d) = \frac{q}{2} (2a + (q - 1)d) \\ p(2a + (p - 1)d) = q(2a + (q - 1)d) \: \: \: \: .....(multiplying \: by \: 2) \: \\ p(2a) + p(p - 1)d - q(2a) - q(q - 1)d = 0 \\ p(2a) - q(2a) + (p(p - 1) - q(q - 1)d = 0 \\ 2a(p - q) + ( {p}^{2} - p - {q}^{2} + q)d = 0 \\ 2a(p - q) + ( {p}^{2} - {q}^{2} - p + q)d = 0 \\ 2a(p - q) + ((p - q)(p + q) - (p - q)) = 0 \\ 2a(p - q) + (p - q)(p + q - 1)d = 0 \\ (p - q)(2a + (p + q - 1)d) = 0 \\ but \: \\ p \: is \: not \: equal \: to \: q \: \: ...(given) \\ (p - q) \: is \: not \: equal \: to \: 0 \\ 2a + (p + q - 1)d = 0 \: \: ....(1) \\ now \\ sp + q \: = \frac{p + q}{2} (2a + (p + q - 1)d) \\ = \frac{p + q}{2} \times 0 \: \: ....(from \: 1) \\ = 0 \\ sp + q \: = 0 \: \\ the \: sum \: of \: first \: (p + q) \: terms \: is \: 0.$

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