Question

In a given A.P, pth term is q and qth term is p , then show that nth term is (p+q-n).
Step by Step

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$




$\textsf{Let the first term is a and common difference is d.}$




$\underline{\textsf{According to question , }} \\ \\ \sf \implies T_p \: = \: a \: + \: ( p \: - \: 1 )d \: = \: q \quad...(1) \\ \\ \textsf{And,} \\ \\ \sf \implies T_q \: = \: a \: + \: ( q \: - \: 1 ) d \: = \: p \quad...(2)$




$\underline{\textsf{Subtract ( 2 ) from ( 1 ),} } \\ \\ \sf \implies \{ a \: + \: ( p \: - \: 1 ) d \} \: - \: \{ a \: + \: ( q \: - \: 1 )d \} \: = \: q \: - \: p \\ \\ \sf \implies \cancel{a} \: + \: pd \: - \: \cancel{ d} \: - \: \cancel{ a } \: - \: qd \: + \: \cancel{d} \: = \: q \: - \: p \\ \\ \sf \implies pd \: - \: qd \: = \: -(p \: - \: q ) \\ \\ \sf \implies \cancel{(p \: - \: q )}d \: = \: - \cancel{(p \: - \: q ) }\\ \\ \sf \: \: \therefore \: \: d \: = \: -1$




$\underline{ \textsf{Plug the value of d in ( 1 ) ,}} \\ \\ \sf \implies a \: + \: ( p \: - \: 1 ) d \: = \: q \\ \\ \sf \implies a \: + \: ( p \: - \: 1 )(-1) \: = \: q \\ \\ \sf \implies a \: - \: p \: + \: 1 \: = \: q \\ \\ \sf \: \: \therefore \: \: a \: = \: p \: + \: q \: - \: 1$



$\underline{\textsf{Now,}} \\ \\ \sf \implies T_n \: = \: a \: + \: ( n \: - \: 1 )d \\ \\ \textsf{Plug the value of a and d , } \\ \\ \sf \implies T_n \: = \: p \: + \: q \: - \: 1 \: + \: ( n \: - \: 1 )(-1) \\ \\ \sf \implies T_n \: = \: p \: + \: q \: - \: \cancel{ 1 }\: - \: n \: + \: \cancel{ 1 } \\ \\ \sf \: \: \: \: \: \therefore \: \: T_n \: = \: p \: + \: q \: - \: n \\ \\ \underline{\underline{\textsf{\Large{Proved !! }} }}$

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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