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# For what value of n , the $n^{th}$ term of the GPs 2560,1280,640 ----- and 10,20,40 ------ are equal.

$\begin{array}{1 1} 40 \\ 4 \\ 5 \\ 160 \end{array}$

Explanation : In the GP 2560,1280,640 ------
$a=2560\;,r=1/2$
$t_{n}=\frac{2560}{2^{n-1}}$
$In\;the\;2^{nd}\;GP\;10,20,40\;----$
$a=10\quad\;r=2$
$t_{n}=10\;2^{n-1}$
$Since\;t_{n}\;is\;same\;for\;n,$
$\frac{2560}{2^{n-1}}=10\;(2)^{n-1}$
$2^{2(n-1)}=256=2^8$
$2(n-1)=8$
$n-1=4$
$n=5.$