Find the n^{th} term of the following series: | vedclass

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(B) The given series is $12, 72, 432, 2592, \dots$This is a Geometric Progression $(GP)$ where the first term $a = 12$ and the common ratio $r = \frac

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$2 	imes 6^{n}$

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(B) The given series is $12, 72, 432, 2592, \dots$This is a Geometric Progression $(GP)$ where the first term $a = 12$ and the common ratio $r = \frac{72}{12} = 6$.The formula for the $n^{th}$ term of a $GP$ is $T_n = a 	imes r^{n-1}$.Substituting the values,we get $T_n = 12 	imes 6^{n-1}$.We can rewrite $12$ as $2 	imes 6$,so $T_n = (2 	imes 6) 	imes 6^{n-1} = 2 	imes 6^{1 + n - 1} = 2 	imes 6^{n}$.Thus,the $n^{th}$ term is $2 	imes 6^{n}$.

Find the $n^{th}$ term of the following series:
$12, 72, 432, 2592, \dots$

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Find the $n^{th}$ term of the following series:$12, 72, 432, 2592, \dots$