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(A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $t_n = a + (n - 1)d$.Given that $p \cdot t_p = q \cdot t_q$.$p[a

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(A) Let the first term be $a$ and the common difference be $d$. The $n^{th}$ term is $t_n = a + (n - 1)d$.Given that $p \cdot t_p = q \cdot t_q$.$p[a + (p - 1)d] = q[a + (q - 1)d]$$ap + p(p - 1)d = aq + q(q - 1)d$$a(p - q) + [p^2 - p - q^2 + q]d = 0$$a(p - q) + [(p^2 - q^2) - (p - q)]d = 0$$a(p - q) + [(p - q)(p + q) - (p - q)]d = 0$Dividing by $(p - q)$ (assuming $p 
eq q$):$a + (p + q - 1)d = 0$Since the $(p + q)^{th}$ term is $t_{p+q} = a + (p + q - 1)d$,we get $t_{p+q} = 0$.

If $p$ times the $p^{th}$ term of an arithmetic progression is equal to $q$ times its $q^{th}$ term,then the $(p + q)^{th}$ term of this progression is........

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