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

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Question

(a) $p\{ a + (p - 1)\,d\} = q\{ a + (q - 1)\,d\} $

$ \Rightarrow $ $a(p - q) + ({p^2} - {q^2})\,d + (q - p)\,d = 0$

$ \Rightarrow $ $(p - q)\{ a

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(a) $p\{ a + (p - 1)\,d\} = q\{ a + (q - 1)\,d\} $

$ \Rightarrow $ $a(p - q) + ({p^2} - {q^2})\,d + (q - p)\,d = 0$

$ \Rightarrow $ $(p - q)\{ a + (p + q - 1)\,d\} = 0$

$ \Rightarrow $ $a + (p + q - 1)\,d = 0$

$ \Rightarrow $${T_{p + q}} = 0$,

$\ \{ \because \;p 
e q\} $ .

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 ${(p + q)^{th}}$ term is

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