Question

Mark the correct alternative in the following question:

\[\text { If in an A . P } . S_n = n^2 q \text { and } S_m = m^2 q, \text { where } S_r \text{ denotes the sum of r terms of the A . P  . , then }S_q \text { equals }\]

पर्याय

• `q^3/2`

• mnq

• `q^3`

• `(m^2+n^2)q`

Answer 1

\[\text { As }, S_n = n^2 q\]

\[ \Rightarrow \frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right] = n^2 q\]

\[ \Rightarrow 2a + \left( n - 1 \right)d = \frac{n^2 q \times 2}{n}\]

\[ \Rightarrow 2a + \left( n - 1 \right)d = 2nq . . . . . \left( i \right)\]

\[\text { Also, } S_m = m^2 q\]

\[ \Rightarrow \frac{m}{2}\left[ 2a + \left( m - 1 \right)d \right] = m^2 q\]

\[ \Rightarrow 2a + \left( m - 1 \right)d = \frac{m^2 q \times 2}{m}\]

\[ \Rightarrow 2a + \left( m - 1 \right)d = 2mq . . . . . \left( ii \right)\]

\[\text { Subtracting } \left( i \right) \text { from } \left( ii \right), \text { we get }\]

\[2a + \left( n - 1 \right)d - 2a - \left( m - 1 \right)d = 2nq - 2mq\]

\[ \Rightarrow \left( n - 1 - m + 1 \right)d = 2q\left( n - m \right)\]

\[ \Rightarrow \left( n - m \right)d = 2q\left( n - m \right)\]

\[ \Rightarrow d = \frac{2q\left( n - m \right)}{\left( n - m \right)}\]

\[ \Rightarrow d = 2q\]

\[\text { Substituting } d = 2q in \left( ii \right),\text {  we get }\]

\[2a + \left( m - 1 \right)2q = 2mq\]

\[ \Rightarrow 2a + 2mq - 2q = 2mq\]

\[ \Rightarrow 2a = 2q\]

\[ \Rightarrow a = q\]

\[\text { Now }, \]

\[ S_q = \frac{q}{2}\left[ 2a + \left( q - 1 \right)d \right]\]

\[ = \frac{q}{2}\left[ 2q + \left( q - 1 \right)2q \right]\]

\[ = \frac{q}{2}\left[ 2q + 2 q^2 - 2q \right]\]

\[ = \frac{q}{2}\left[ 2 q^2 \right]\]

\[ = q^3\]

Hence, the correct alternative is option (c).