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Question
In a GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.
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Solution
\[\text { Let a be the first term and r be the common ratio } . \]
\[ \therefore a_3 = 24 \text { and } a_6 = 192\]
\[ \Rightarrow a r^2 = 24 \text { and } a r^5 = 192\]
\[ \Rightarrow \frac{a r^5}{a r^2} = \frac{192}{24}\]
\[ \Rightarrow r^3 = 8 \]
\[ \Rightarrow r^3 = 2^3 \]
\[ \Rightarrow r = 2\]
\[\text { Putting } r = 2 \text { in a }r^2 = 24\]
\[a \left( 2 \right)^2 = 24 \]
\[ \Rightarrow a = 6\]
\[\text { Now}, {10}^{th}\text { term }= a_{10} = a r^9 \]
\[\text { Putting a = 6 and r = 2 in } a_{10} = a r^9 \]
\[ \Rightarrow a_{10} = \left( 6 \right) \left( 2 \right)^9 = 3072\]
\[\text { Thus, the } {10}^{th}\text { term of the G.P. is } 3072 .\]
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