What term of the progression 18, -12, 8, ldot | vedclass

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(C) The given progression is a geometric progression where the first term $a = 18$ and the common ratio $r = \frac{-12}{18} = \frac{-2}{3}$.Let the $n

Vedclass · Accepted Answer

$9$

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(C) The given progression is a geometric progression where the first term $a = 18$ and the common ratio $r = \frac{-12}{18} = \frac{-2}{3}$.Let the $n^{th}$ term be $a_n = \frac{512}{729}$.The formula for the $n^{th}$ term of a geometric progression is $a_n = a r^{n-1}$.Substituting the values,we get $\frac{512}{729} = 18 \left( \frac{-2}{3} ight)^{n-1}$.Dividing both sides by $18$,we have $\left( \frac{-2}{3} ight)^{n-1} = \frac{512}{729 	imes 18} = \frac{256}{729 	imes 9} = \frac{2^8}{3^6 	imes 3^2} = \frac{2^8}{3^8} = \left( \frac{2}{3} ight)^8$.Since the base is negative and the exponent is even,$\left( \frac{-2}{3} ight)^8 = \left( \frac{2}{3} ight)^8$.Thus,$\left( \frac{-2}{3} ight)^{n-1} = \left( \frac{-2}{3} ight)^8$.Comparing the exponents,$n - 1 = 8$,which gives $n = 9$.Therefore,$\frac{512}{729}$ is the $9^{th}$ term of the progression.

What term of the progression $18, -12, 8, \ldots$ is $\frac{512}{729}$?

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