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8. Sequences and Series
easy
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
A
$0$
B
$1$
C
$2$
D
$3$
Solution
(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 \ne q\} $ .
Standard 11
Mathematics
