Question

In an given AP,if pth term q and qth term is p, then show that the nth term is p+q-n​

Answer 1

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Here,

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

Answer 2

Answer:

pth term = q

a+(p−1)d=q

qth term = p

a+(q−1)d=p

Solving these equations, we get,

d=−1

a=(p+q−1)

Thus,

nth term = a+(n−1)d=(p+q−1)+(n−1)×(−1)=(p+q−n)

Explanation:

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