Question

if the pth term of an AP is a and th term of an AP is p ,prove that nth term is (p+q-n)

Answer 1

Question:

If the pth term of an AP is q and qth term of the AP is p, prove that nth term is (p+q-n).

Proof:

Let the first term and common difference be a and d respectively.

$T_p-T_q=q+p \\ \\ (p-q)d=q-p \\ \\ d=\frac{q-p}{p-q} \\ \\ d=\frac{-(p-q)}{p-q} \\ \\ d=-1$

$T_p=a+(p-1)d \\ \\ T_p=a-1(p-1) \\ \\ T_p=a-(p-1) \\ \\ T_p=a-p+1$

$T_q=a+(q-1)d \\ \\ T_q=a-1(q-1) \\ \\ T_q=a-(q-1) \\ \\ T_q=a-q+1$

$T_p+T_q=q+p \\ \\ a-p+1+a-q+1=q+p \\ \\2a-p-q+2=q+p \\ \\ 2a+2=2p+2q \\ \\ 2(a+1)=2(p+q) \\ \\ a+1=p+q$

$T_n=a+(n-1)d \\ \\ T_n=a-1(n-1) \\ \\ T_n=a-n+1 \\ \\ T_n=a+1-n \\ \\ T_n=p+q-n$

Hence proved!!!

     

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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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