Question

If the pth term of an A.P. is q and the qth term of an A.P. is p, show that its nth term is (p+q-n).

Answer 1

ap = q
a+(p-1)d =q
a+dp - d=q.... 1
aq=p
a+(q-1)d=p
a+dq - d=p.... 2
Subtracting 1 n 2
d=-1
Now putting the value of d in eq -1
a+(p-1)-1=q
a-p+1=q
a=p+q-1
Now nth term = a+(n-1)d
p+q-1+(n-1)-1
=p+q-1-n+1
p+q-n (proved)

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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