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# Which term of the series $\frac{16}{9}$, $\frac{4}{3}$, $1$, $\frac{3}{4}$------is $\frac{243}{1024}$

$\begin{array}{1 1} 5th \\ 7th \\ 9th \\ 3rd \end{array}$

Answer : (b) $7^{th}$
Explanation : The series is a GP with $a=16/9\;,r=3/4$
$ar^n=\large\frac{243}{1024}=\quad\;r^n=\large\frac{243*9}{1024*16}=(3/4)^7$
$n=7.$
edited Jan 24, 2014