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(B) Let the first term be $a$ and the common difference be $d$.Given,$t_p = a + (p - 1)d = q$  --- $(1)$And,$t_q = a + (q - 1)d = p$  --- $(2)$Subtrac

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$p + q - r$

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(B) Let the first term be $a$ and the common difference be $d$.Given,$t_p = a + (p - 1)d = q$  --- $(1)$And,$t_q = a + (q - 1)d = p$  --- $(2)$Subtracting $(2)$ from $(1)$:$(p - q)d = q - p$$(p - q)d = -(p - q)$$d = -1$Substituting $d = -1$ in $(1)$:$a + (p - 1)(-1) = q$$a - p + 1 = q$$a = p + q - 1$Now,the $r^{th}$ term $t_r = a + (r - 1)d$$t_r = (p + q - 1) + (r - 1)(-1)$$t_r = p + q - 1 - r + 1$$t_r = p + q - r$

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

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