Question

If the pth term of an ap is q ans qth term is p then prove that an=p+q-n


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Answer 1

Given :-

• Ap = q
• Aq = p

To Prove :- An = (p+q-n)

Concepts :-

• Let The First Term Be a
• Let The Common Differnce be d

A. T. Q.

Ap = a+(p-1)d

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

Also,

Aq = a+(q-1)d

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

Subtracting Eqn (ii) From eqn (i) we Get,

p-q= (q-p)d

=> d= -1

Substituting d = -1 in eqn (i)

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

=> a = p+q-1

Now, An = a+(n-1)d And a = (p+q-1), d= -1

=> An = (p+q-1) + (n-1)(-1)

Therfore, An = p+q-1-n+1

=> p+q-n

So, An = (p+q-n)

Hope It Helps You..... :)

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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