Question

If the Pth term of an AP is q and the qth term is p prove that it's nth term is (p + q - n)th term of the AP is zero

Answer 1

HI FRIENDS
here is your solution.

let a & d are the first term & common differnece respectively then

pth term = a+(p-1)d = q ...............1

qth term = a+(q-1)d = p ...............2

1-2

q-p = (p-q)d

d=-1

now

(p+q)th term = a + (p+q-1)d

putting value of a from eq 1
(p+q)th term = q - (p-1)d + (p+q-1)d (d =-1)

=0

Answer 2

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$let \: a \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: nth \: term \: of \: ap \\ t _{p} = a + (p - 1)d \: \: and \: t _{q} = a + (q - 1)d \\ now \: t _{p } = q \: and \: t _{q} = p \\ \therefore \: a + (p - 1)d = q \: \: \: \: \: \: \: \: \: \: \: \: ... .(1) \\ and \: a + (q - 1)d = p \: \: \: \: \: \: \: \: \: \: \: \: \: .. .. (2) \\ \\ \\ on \: subtracting \: (1)from(2) \: we \: get \\ (q - p)d = (p - q) \implies \: d = - 1 \\ putting \: d = - 1 \: in \: (1) \: we \: get \: a = (p + q - 1) \\ \therefore \: nth \: term \: = a(n - 1)d = (p + q - 1) + (n - 1)( - 1) = (p + q - n) \\ \\ hence \: nth \: term \: = (p + q - n)$

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