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Question
Mark the correct alternative in the following question:
\[\text { If in an A . P } . S_n = n^2 q \text { and } S_m = m^2 q, \text { where } S_r \text{ denotes the sum of r terms of the A . P . , then }S_q \text { equals }\]
Options
`q^3/2`
mnq
`q^3`
`(m^2+n^2)q`
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Solution
\[\text { As }, S_n = n^2 q\]
\[ \Rightarrow \frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right] = n^2 q\]
\[ \Rightarrow 2a + \left( n - 1 \right)d = \frac{n^2 q \times 2}{n}\]
\[ \Rightarrow 2a + \left( n - 1 \right)d = 2nq . . . . . \left( i \right)\]
\[\text { Also, } S_m = m^2 q\]
\[ \Rightarrow \frac{m}{2}\left[ 2a + \left( m - 1 \right)d \right] = m^2 q\]
\[ \Rightarrow 2a + \left( m - 1 \right)d = \frac{m^2 q \times 2}{m}\]
\[ \Rightarrow 2a + \left( m - 1 \right)d = 2mq . . . . . \left( ii \right)\]
\[\text { Subtracting } \left( i \right) \text { from } \left( ii \right), \text { we get }\]
\[2a + \left( n - 1 \right)d - 2a - \left( m - 1 \right)d = 2nq - 2mq\]
\[ \Rightarrow \left( n - 1 - m + 1 \right)d = 2q\left( n - m \right)\]
\[ \Rightarrow \left( n - m \right)d = 2q\left( n - m \right)\]
\[ \Rightarrow d = \frac{2q\left( n - m \right)}{\left( n - m \right)}\]
\[ \Rightarrow d = 2q\]
\[\text { Substituting } d = 2q in \left( ii \right),\text { we get }\]
\[2a + \left( m - 1 \right)2q = 2mq\]
\[ \Rightarrow 2a + 2mq - 2q = 2mq\]
\[ \Rightarrow 2a = 2q\]
\[ \Rightarrow a = q\]
\[\text { Now }, \]
\[ S_q = \frac{q}{2}\left[ 2a + \left( q - 1 \right)d \right]\]
\[ = \frac{q}{2}\left[ 2q + \left( q - 1 \right)2q \right]\]
\[ = \frac{q}{2}\left[ 2q + 2 q^2 - 2q \right]\]
\[ = \frac{q}{2}\left[ 2 q^2 \right]\]
\[ = q^3\]
Hence, the correct alternative is option (c).
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