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If (P + Q)Th and (P − Q)Th Terms of a G.P. Are M and N Respectively, Then Write is Pth Term. - Mathematics

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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.

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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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Chapter 20: Geometric Progression - Exercise 20.7 [Page 56]

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RD Sharma Mathematics [English] Class 11
Chapter 20 Geometric Progression
Exercise 20.7 | Q 2 | Page 56

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