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In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.

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प्रश्न

In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.

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उत्तर

\[\text { Let a be the first term and r be the common ratio } . \]

\[ \therefore a_3 = 24 \text { and } a_6 = 192\]

\[ \Rightarrow a r^2 = 24 \text { and } a r^5 = 192\]

\[ \Rightarrow \frac{a r^5}{a r^2} = \frac{192}{24}\]

\[ \Rightarrow r^3 = 8 \]

\[ \Rightarrow r^3 = 2^3 \]

\[ \Rightarrow r = 2\]

\[\text { Putting } r = 2 \text { in a }r^2 = 24\]

\[a \left( 2 \right)^2 = 24 \]

\[ \Rightarrow a = 6\]

\[\text { Now}, {10}^{th}\text { term  }= a_{10} = a r^9 \]

\[\text { Putting a = 6 and r = 2 in } a_{10} = a r^9 \]

\[ \Rightarrow a_{10} = \left( 6 \right) \left( 2 \right)^9 = 3072\]

\[\text { Thus, the } {10}^{th}\text { term of the G.P. is } 3072 .\]

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

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 20 Geometric Progression
Exercise 20.1 | Q 14 | पृष्ठ १०
नूतन Mathematics [English] Class 10 ICSE
अध्याय 9 Arithmetic and geometric progression
CHAPTER TEST | Q 5. | पृष्ठ २०२

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