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# Find n such that is $\large\frac{512}{729}\;$is the nth term of the GP 18,-12,8,-----------------

$\begin{array}{1 1} 8 \\ 18 \\ 12 \\ 9 \end{array}$

Explanation :$a=18 , r=\large\frac{-12}{18}=\large\frac{-2}{3}$
$T_{n}=ar^{n-1}\qquad\;\frac{512}{729}=18*(-2/3)^{(n-1)}$
$(-2/3)^{(n-1)}=\large\frac{512}{18*729}=\large\frac{2^9}{2*3^2*9^3}=\large\frac{2^8}{3^2*3^6}$
$=\large\frac{2^8}{3^8}$
$n-1=8\quad\;n=9.$
edited Jan 24, 2014