Question

for am AP the pth term is q and the qth term is p. find the general term of the AP

Answer 1

Answer:

Given that, pth term of an AP is q and the qth term of an AP is p.

To find the General term of the AP.

According to the question,

$\implies a_p = a + ( p - 1 )d\\\\\implies q = a + ( p - 1 )d \:\:\:...(1)$

$\implies a_q = a + ( q - 1 )d\\\\\implies p = a + ( q - 1 ) d \:\:\:...(2)$

Subtracting ( 2 ) from ( 1 ) we get,

⇒ q - p = a - a + ( p - 1 )d - ( q - 1 )d

⇒ q - p = pd - d - [ qd - d ]

⇒ q - p = pd - d - qd + d

⇒ q - p = pd - qd

⇒ q - p = ( p - q ) d

We can also write ( q - p ) as -1 ( p - q ). Writing like this we get,

⇒ -1 ( p - q ) = d ( p - q )

Cancelling the ( p - q ) terms we get,

⇒ d = ( -1 )

Substituting the value of d in any of the equations we get,

Substituting in (1),

⇒ q = a + ( p - 1 ) ( - 1 )

⇒ q = a + ( -p + 1 )

⇒ q = a - p + 1

⇒ a = q + p - 1

Hence we know that general term of an AP is:

$\implies a_n = a + ( n - 1 )d\\\\\text{Substituting the values we get,}\\\\\implies a_n = ( p + q - 1 ) + ( n - 1 ) ( - 1 )\\\\\implies a_n = ( p + q - 1 ) + ( -n + 1 )\\\\\implies a_n = p + q - 1 - n + 1 \\\\\implies a_n = p + q - n$

This is the required answer.

Answer 2

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We have

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

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