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(A) Let the first term be $a$ and the common ratio be $r$.According to the problem,the $n^{th}$ term of a geometric progression is given by $T_n = ar^

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$6$

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(A) Let the first term be $a$ and the common ratio be $r$.According to the problem,the $n^{th}$ term of a geometric progression is given by $T_n = ar^{n-1}$.Given: $T_{10} = ar^9 = 9$ and $T_4 = ar^3 = 4$.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,$T_7 = ar^6$.From $ar^3 = 4$,we have $a = \frac{4}{r^3}$.Thus,$T_7 = \left(\frac{4}{r^3}ight)r^6 = 4r^3$.Since $r^6 = \frac{9}{4}$,then $r^3 = \sqrt{\frac{9}{4}} = \frac{3}{2}$.Therefore,$T_7 = 4 	imes \frac{3}{2} = 6$.Alternatively,for a geometric progression,the $7^{th}$ term is the geometric mean of the $4^{th}$ and $10^{th}$ terms because $7$ is the arithmetic mean of $4$ and $10$: $T_7 = \sqrt{T_4 	imes T_{10}} = \sqrt{4 	imes 9} = \sqrt{36} = 6$.

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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