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Question
If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
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Solution
\[a_4 = x\]
\[ \Rightarrow a r^3 = x\]
\[\text { Also }, a_{10} = y\]
\[ \Rightarrow a r^9 = y\]
\[\text { And, } a_{16} = z\]
\[ \Rightarrow a r^{15} = z\]
\[ \because \frac{y}{x} = \frac{a r^9}{a r^3} = r^6 \]
\[\text { and } \frac{z}{y} = \frac{a r^{15}}{a r^9} = r^6 \]
\[ \therefore \frac{y}{x} = \frac{z}{y}\]
\[\text { Therefore, x, y and z are in G . P } .\]
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