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(A) Let the first term be $a$ and the common ratio be $r$.Given that the $10^{th}$ term $a_{10} = ar^9 = 9$ and the $4^{th}$ term $a_4 = ar^3 = 4$.Div

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$6$

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(A) Let the first term be $a$ and the common ratio be $r$.Given that the $10^{th}$ term $a_{10} = ar^9 = 9$ and the $4^{th}$ term $a_4 = ar^3 = 4$.Dividing the two equations: $\frac{ar^9}{ar^3} = \frac{9}{4} \Rightarrow r^6 = \frac{9}{4}$.We need to find the $7^{th}$ term,$a_7 = ar^6$.Since $ar^9 = 9$,we have $a = \frac{9}{r^9}$.Alternatively,using the property of geometric progression: $a_7 = \sqrt{a_{10} 	imes a_4} = \sqrt{9 	imes 4} = \sqrt{36} = 6$.Thus,the $7^{th}$ term is $6$.

If the $10^{th}$ term of a geometric progression is $9$ and the $4^{th}$ term is $4$,then its $7^{th}$ term is

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