Question

If pth term of an Ap is q and the qth term is p than show that rth term p+q-r

Answer 1

ap = a + (p-1)d 
q = a + pd -d ...........1

aq = a + (q-1)d
p = a + qd - d............2

subtract 2 from 1 we get
q - p = a + pd -d - a - qd+ d
q - p = pd - qd
-1(p - q) = d(p - q)
d = -1

put value of d in 1 equation
q = a - p +1
a = p + q - 1

ar = a+ (r - 1) d
put value of a and d 
    = p + q - 1 - r +1
    = p + q - r

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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