Question

if pth term of an AP is q and qth term is p,then show that it's nth term is (p+q-n)

Answer 1

We are given that,
a+(p−1)d=q        −−−−−−−−−−(1)
a+(q−1)d=p        −−−−−−−−−−(2)

Subtracting (1) from (2), we get,

(p−q)=a+(q−1)d−[a+(p−1)d]
(p−q)=a+(q−1)d−a−(p−1)d
(p−q)=(q−1)d−(p−1)d
(p−q)=(q−1−p+1)d
(p−q)=(q−p)d
(−q+p)=(q−p)d
−(q−p)=(q−p)d
−1=d

From (2) we get,
a=p−(q−1)d
a=p−(q−1)(−1)    {Because d=−1}
a=p+q−1

Now we have both first term (a) and common difference (d), we can solve for nth term,

an=a+(n−1)d
i.e. an=p+q−1+(n−1)(−1)
      an=p+q−1−n+1
    
an=(p+q−n)

Answer 2

${\green {\boxed {\mathtt {✓verified\:answer}}}}$

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