If p times the p^{th} term of an A.P. is equa | vedclass

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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: $p \cdot T_p = q \cdot T_q$$p\{a + (p

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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: $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$$a(p - q) + d\{p^2 - p - q^2 + q\} = 0$$a(p - q) + d\{(p^2 - q^2) - (p - q)\} = 0$$a(p - q) + d\{(p - q)(p + q) - (p - q)\} = 0$Since $p 
eq q$,we can divide by $(p - q)$:$a + d(p + q - 1) = 0$This expression represents the $(p + q)^{th}$ term,$T_{p+q} = a + (p + q - 1)d$.Therefore,$T_{p+q} = 0$.

If $p$ times the $p^{th}$ term of an $A.P.$ is equal to $q$ times the $q^{th}$ term of an $A.P.$,then the $(p + q)^{th}$ term is

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