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If the Pth and Qth Terms of a G.P. Are Q and P, Respectively, Then Show that (P + Q)Th Term is ( Q P P Q ) 1 P − Q . - Mathematics

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प्रश्न

If the pth and qth terms of a G.P. are q and p, respectively, then show that (p + q)th term is \[\left( \frac{q^p}{p^q} \right)^\frac{1}{p - q}\].

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उत्तर

\[\text { As, } a_p = q\]

\[ \Rightarrow a r^\left( p - 1 \right) = q . . . . . \left( i \right)\]

\[\text { Also, } a_q = p\]

\[ \Rightarrow a r^\left( q - 1 \right) = p . . . . . \left( ii \right)\]

\[\text { Dividing  }\left( i \right)\text {  by } \left( ii \right), \text { we get }\]

\[\frac{a r^\left( p - 1 \right)}{a r^\left( q - 1 \right)} = \frac{q}{p}\]

\[ \Rightarrow r^\left( p - 1 - q + 1 \right) = \frac{q}{p}\]

\[ \Rightarrow r^\left( p - q \right) = \frac{q}{p}\]

\[ \Rightarrow r = \left( \frac{q}{p} \right)^\frac{1}{\left( p - q \right)} \]

\[\text { Substituting the value of r in } \left( ii \right), \text { we get }\]

\[a \left[ \left( \frac{q}{p} \right)^\frac{1}{\left( p - q \right)} \right]^\left( q - 1 \right) = p\]

\[ \Rightarrow a\left[ \left( \frac{q}{p} \right)^\frac{\left( q - 1 \right)}{\left( p - q \right)} \right] = p\]

\[ \Rightarrow a = p \times \left( \frac{p}{q} \right)^\frac{\left( q - 1 \right)}{\left( p - q \right)} \]

\[ \Rightarrow a = p \left( \frac{p}{q} \right)^\frac{\left( q - 1 \right)}{\left( p - q \right)} \]

\[\text { Now, } \]

\[ a_\left( p + q \right) = a r^\left( p + q - 1 \right) \]

\[ = p \left( \frac{p}{q} \right)^\frac{\left( q - 1 \right)}{\left( p - q \right)} \times \left[ \left( \frac{q}{p} \right)^\frac{1}{\left( p - q \right)} \right]^\left( p + q - 1 \right) \]

\[ = p \left( \frac{p}{q} \right)^\frac{\left( q - 1 \right)}{\left( p - q \right)} \times \left( \frac{q}{p} \right)^\frac{\left( p + q - 1 \right)}{\left( p - q \right)} \]

\[ = p \left( \frac{q}{p} \right)^\frac{- \left( q - 1 \right)}{\left( p - q \right)} \times \left( \frac{q}{p} \right)^\frac{\left( p + q - 1 \right)}{\left( p - q \right)} \]

\[ = p \times \left( \frac{q}{p} \right)^\frac{- \left( q - 1 \right)}{\left( p - q \right)} + \frac{\left( p + q - 1 \right)}{\left( p - q \right)} \]

\[ = p \times \left( \frac{q}{p} \right)^\frac{- q + 1 + p + q - 1}{\left( p - q \right)} \]

\[ = p \times \left( \frac{q}{p} \right)^\frac{p}{\left( p - q \right)} \]

\[ = \frac{p \times q^\frac{p}{\left( p - q \right)}}{p^\frac{p}{\left( p - q \right)}}\]

\[ = \frac{q^\frac{p}{\left( p - q \right)}}{p^\frac{p}{\left( p - q \right)} - 1}\]

\[ = \frac{q^\frac{p}{\left( p - q \right)}}{p^\frac{p - p + q}{\left( p - q \right)}}\]

\[ = \frac{q^\frac{p}{\left( p - q \right)}}{p^\frac{q}{\left( p - q \right)}}\]

\[ = \frac{q^{p \times \frac{1}{\left( p - q \right)}}}{p^{q \times \frac{1}{\left( p - q \right)}}}\]

\[ = \left( \frac{q^p}{p^q} \right)^\frac{1}{p - q}\]

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अध्याय 20: Geometric Progression - Exercise 20.1 [पृष्ठ १०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.1 | Q 17 | पृष्ठ १०

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