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(A) Let the first term be $a$ and the common ratio be $r$ for the geometric progression.Given: $a_{10} = ar^9 = 9$ and $a_4 = ar^3 = 4$.Dividing the t

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$6$

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(A) Let the first term be $a$ and the common ratio be $r$ for the geometric progression.Given: $a_{10} = ar^9 = 9$ and $a_4 = ar^3 = 4$.Dividing the two equations: $\frac{ar^9}{ar^3} = \frac{9}{4} \implies r^6 = \frac{9}{4}$.The $7^{th}$ term is $a_7 = ar^6$.Since $ar^3 = 4$,we have $a = \frac{4}{r^3}$.From $r^6 = \frac{9}{4}$,we get $r^3 = \sqrt{\frac{9}{4}} = \frac{3}{2}$ (assuming $r > 0$).Thus,$a = \frac{4}{3/2} = \frac{8}{3}$.Therefore,$a_7 = ar^6 = \left(\frac{8}{3}ight) 	imes \left(\frac{9}{4}ight) = 2 	imes 3 = 6$.

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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