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 $S_n = 189$.In a geometric progression,the $n^{th}$ term is given by $a_n = a_1 r^{n-1}$.Substituting the values:

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$6$

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(B) Given: $a_1 = 3$,$a_n = 96$,and $S_n = 189$.In a geometric progression,the $n^{th}$ term is given by $a_n = a_1 r^{n-1}$.Substituting the values: $3 r^{n-1} = 96 \Rightarrow r^{n-1} = 32$ ..... $(i)$.The sum of $n$ terms is $S_n = \frac{a_1(r^n - 1)}{r - 1} = 189$.We can write $S_n = \frac{a_1 r^n - a_1}{r - 1} = \frac{r(a_1 r^{n-1}) - a_1}{r - 1} = 189$.Substituting $a_1 r^{n-1} = 96$ and $a_1 = 3$: $\frac{r(96) - 3}{r - 1} = 189$.$96r - 3 = 189(r - 1) \Rightarrow 96r - 3 = 189r - 189$.$186 = 93r \Rightarrow r = 2$.From $(i)$,$2^{n-1} = 32 = 2^5$.Therefore,$n - 1 = 5 \Rightarrow n = 6$.

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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