Question

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

Answer 1

Answer:

Step-by-step explanation:

Let first term be a and common difference be d.

pth term = a + (p-1)d = q -----(i)

qth term = a + (q-1)d = p -----(ii)

(ii) - (i) gives

(q - p)d = p - q or d = -1

(i) gives  a = q - (p-1)d = q - (p-1)(-1) = q + p - 1

Hence,

nth term = a + (n-1)d = q + p -1 + (n-1)(-1) = q + p -1 + n +1 = p + q - n

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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