Question

If the pth term of an AP is q and the qth term is p, show that its n-th term is (p+q-n)

Answer 1

Given :-

• p th term of AP is q
• q th term of AP is p

To prove :-

• nth term of AP ( aₙ ) is ( p  + q - n )

Proof :-

As given ,

ᅠ

$\to\sf{a_p=a+(p-1)d=q\;\;\;....eqn(1)}$

and

$\to\sf{a_q=a+(q-1)d=p\;\;\;...eqn(2)}$

ᅠ

subtracting eqn (2) from eqn (1)

ᅠ

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

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

→ d ( p - 1 - q + 1 ) = - ( p - q )

→ d ( p - q ) = - ( p - q )

→ d = - 1

ᅠ

Putting value of d in eqn (1)

ᅠ

→ a + ( p - 1 ) ( -1 ) = q

→ a + 1 - p = q

→ a = p + q - 1

ᅠ

Now,

ᅠ

→ aₙ = a + ( n - 1 ) d

→ aₙ = ( p + q - 1 ) + ( n - 1 ) ( -1 )

→ aₙ = p + q - 1 - n + 1

→ aₙ = p + q - n

ᅠ

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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