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

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(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.According

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(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $T_n = a + (n - 1)d$.According to the problem,$p \cdot T_p = q \cdot T_q$.Substituting the formula: $p\{a + (p - 1)d\} = q\{a + (q - 1)d\}$.Expanding the terms: $ap + p(p - 1)d = aq + q(q - 1)d$.Rearranging the terms: $a(p - q) + \{p(p - 1) - q(q - 1)\}d = 0$.$a(p - q) + \{p^2 - p - q^2 + q\}d = 0$.$a(p - q) + \{(p^2 - q^2) - (p - q)\}d = 0$.$a(p - q) + \{(p - q)(p + q) - (p - q)\}d = 0$.Dividing by $(p - q)$ (since $p 
eq q$): $a + (p + q - 1)d = 0$.Since $T_{p+q} = a + (p + q - 1)d$,we conclude that $T_{p+q} = 0$.

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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