Question

The pth term of an AP is q and the qth term is p. Show that the rth term is p+q-r. Which term of the series is zero. ​

Answer 1

Answer:

Step-by-step explanation:

Answer:

Hence Proved, rth term of the AP is p+q-r

Step-by-step explanation:

If the pth term of an AP is q and the qth term is p

Let first term of AP is a and common difference is d

Solve for a and d

Put d=-1 into

We need to find rth term of same AP where a=p+q-1 and d=-1

Hence proved, rth term of the AP is p+q-r

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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