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Question
If (p + q)th and (p − q)th terms of a G.P. are m and n respectively, then write is pth term.
Solution
\[\text { Here }, \left( p + q \right)^{th} \text { term } = m \]
\[ \Rightarrow a r^\left( p + q \right) - 1 = m . . . . . . . \left( i \right)\]
\[\text { And }, \left( p - q \right)^{th}\text { term } = n \]
\[ \Rightarrow a r^\left( p - q \right) - 1 = n . . . . . . . \left( ii \right)\]
\[\text { Dividing } \left( i \right) \text { by } \left( ii \right): \]
\[\frac{a r^\left( p + q \right) - 1}{a r^\left( p - q \right) - 1} = \frac{m}{n} \]
\[ \Rightarrow r^{2q} = \frac{m}{n}\]
\[ \Rightarrow r^q = \sqrt{\frac{m}{n}}\]
\[\text { Now, from } \left( i \right): \]
\[a\left( r^{p - 1} \times r^q \right) = m\]
\[ \Rightarrow a r^{p - 1} \times \sqrt{\frac{m}{n}} = m\]
\[ \Rightarrow a r^{p - 1} = m \times \frac{\sqrt{n}}{\sqrt{m}}\]
\[ \Rightarrow a r^{p - 1} = \frac{m\sqrt{n}}{\sqrt{m}}\]
\[\text { Thus, the } p^{th}\text { term is } \frac{m\sqrt{n}}{\sqrt{m}} .\]
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