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(B) Let the first term be $a$ and the common difference be $d$.Given that the $p^{th}$ term is $q$,we have: $a + (p - 1)d = q$  $(1)$Given that the $q

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$p + q - n$

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(B) Let the first term be $a$ and the common difference be $d$.Given that the $p^{th}$ term is $q$,we have: $a + (p - 1)d = q$  $(1)$Given that the $q^{th}$ term is $p$,we have: $a + (q - 1)d = p$  $(2)$Subtracting equation $(2)$ from $(1)$:$(p - q)d = q - p$$d = \frac{q - p}{p - q} = -1$Substituting $d = -1$ into equation $(1)$:$a + (p - 1)(-1) = q$$a - p + 1 = q$$a = p + q - 1$The $n^{th}$ term $T_n$ is given by:$T_n = a + (n - 1)d$$T_n = (p + q - 1) + (n - 1)(-1)$$T_n = p + q - 1 - n + 1$$T_n = p + q - n$

If the $p^{th}$ term of an arithmetic progression is $q$ and the $q^{th}$ term is $p$,then its $n^{th}$ term is:

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