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Question
Mark the correct alternative in the following question:
If in an A.P., the pth term is q and (p + q)th term is zero, then the qth term is
Options
\[-\]p
p
p + q
p-q
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Solution
\[\text { As, } a_p = q\]
\[ \Rightarrow a + \left( p - 1 \right)d = q . . . . . \left( i \right)\]
\[\text { Also }, a_\left( p + q \right) = 0\]
\[ \Rightarrow a + \left( p + q - 1 \right)d = 0 . . . . . \left( ii \right)\]
\[\text { Subtracting } \left( i \right) \text { from } \left( ii \right), \text { we get }\]
\[a + \left( p + q - 1 \right)d - a - \left( p - 1 \right)d = 0 - q\]
\[ \Rightarrow \left( p + q - 1 - p + 1 \right)d = - q\]
\[ \Rightarrow qd = - q\]
\[ \Rightarrow d = \frac{- q}{q}\]
\[ \Rightarrow d = - 1\]
\[\text { Substituting } d = - 1 \text { in } \left( i \right), \text { we get }\]
\[a + \left( p - 1 \right) \times \left( - 1 \right) = q\]
\[ \Rightarrow a - p + 1 = q\]
\[ \Rightarrow a = p + q - 1\]
\[\text { Now }, \]
\[ a_q = a + \left( q - 1 \right)d\]
\[ = p + q - 1 + \left( q - 1 \right) \times \left( - 1 \right)\]
\[ = p + q - 1 - q + 1\]
\[ = p\]
Hence, the correct alternative is option (b).
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