Question

pth term of an A. P is q and qth term of an A. P is p prove that nth term is p+q-n

Answer 1

Hey there !!


Let a be the first term and d be the common difference of the given AP.

Then,

→ T$\tiny p$ = a + ( p - 1 )d .

And,

→ T$\tiny q$ = a + ( q - 1 )d.


Now,

→ T$\tiny p$ = q .

And,

T$\tiny q$ = p ( given ) .

=> a + ( p - 1 )d = q ....... (1).

And, a + ( q - 1 )d = p ..........(2).


▶ On substracting equation (1) and (2), we get

a + ( p - 1 )d = q.
a + ( q - 1 )d = p.
(-)....(-)............(-)
____________

=> ( q - p )d = ( p - q ).

=> ( q - p )d = -( q - p ).

=> d = -1 .

Putting d = -1 in equation (1), we get

=> a + ( p - 1 )d = q.

=> a + ( p - 1 ) (-1) = q.

=> a - p + 1 = q .

=> a = p + q - 1.

Thus,

→ a = ( p + q - 1 ) and d = -1 .

→ nth term = a + ( n - 1 )d.

= p + q - 1 + ( n - 1 ) (-1) .

= p + q - 1 - n + 1.

$\huge \boxed{ \bf = p + q - n . }$


✔✔ Hence, nth term = ( p + q - n ). ✅✅

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THANKS


#BeBrainly.

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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