If in a geometric progression {a_n},a_1 = 3,a | vedclass

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(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.We know that $a_n = a_1 r^{n-1}$,so $3 r^{n-1} = 96$,which implies $r^{n-1} = 32$.Thus,$r^n = 32r$.The

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$6$

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(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.We know that $a_n = a_1 r^{n-1}$,so $3 r^{n-1} = 96$,which implies $r^{n-1} = 32$.Thus,$r^n = 32r$.The sum of $n$ terms of a geometric progression is $S_n = \frac{a_1(r^n - 1)}{r - 1} = 189$.Substituting the values: $\frac{3(32r - 1)}{r - 1} = 189$.Dividing by $3$: $\frac{32r - 1}{r - 1} = 63$.$32r - 1 = 63r - 63$.$62 = 31r$,so $r = 2$.Since $r^{n-1} = 32$,we have $2^{n-1} = 2^5$.Therefore,$n - 1 = 5$,which gives $n = 6$.

If in a geometric progression $\{a_n\}$,$a_1 = 3$,$a_n = 96$ and $S_n = 189$,then the value of $n$ is

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