If the 2^{text{nd}} and 5^{text{th}} terms of | vedclass

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(A) Given that,$T_{2} = 24$ and $T_{5} = 3$.In a $G$.$P$.,the $n^{	ext{th}}$ term is given by $T_{n} = ar^{n-1}$.So,$ar = 24$ $(1)$ and $ar^{4} = 3$

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$ \frac{189}{2} $

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(A) Given that,$T_{2} = 24$ and $T_{5} = 3$.In a $G$.$P$.,the $n^{	ext{th}}$ term is given by $T_{n} = ar^{n-1}$.So,$ar = 24$ $(1)$ and $ar^{4} = 3$ $(2)$.Dividing equation $(2)$ by equation $(1)$,we get $\frac{ar^{4}}{ar} = \frac{3}{24}$ $\Rightarrow r^{3} = \frac{1}{8}$ $\Rightarrow r = \frac{1}{2}$.Substituting $r = \frac{1}{2}$ in equation $(1)$,we get $a 	imes \frac{1}{2} = 24 \Rightarrow a = 48$.The sum of the first $n$ terms of a $G$.$P$. is $S_{n} = \frac{a(1-r^{n})}{1-r}$.For $n = 6$,$S_{6} = \frac{48(1-(1/2)^{6})}{1-1/2} = \frac{48(1-1/64)}{1/2} = 96 	imes \frac{63}{64} = \frac{3 	imes 63}{2} = \frac{189}{2}$.

If the $2^{\text{nd}}$ and $5^{\text{th}}$ terms of a $G$.$P$. are $24$ and $3$ respectively,then the sum of the $1^{\text{st}}$ six terms is:

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