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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series

Which term of the sequence $\sqrt 3,3,3\sqrt 3.......$ is $729?$

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  • $n^{th}$ term of a G.P$=t_n=a.r^{n-1}$ where $a=1^{st}\:term\:and\:r=common\:ratio.$
The given sequence is $\sqrt 3,3,3\sqrt 3.........$
This sequence is a G.P.since the ratio of any two successive terms is same.
$1^{st}\:term$ of the G.P.$=a=\sqrt 3$ and common ratio$=r=\large\frac{3}{\sqrt 3}$$=\sqrt 3$
We know that $t_n=a.r^{n-1}$
Also given that $t_n=729$
$\Rightarrow\:729=a.r^{n-1}=\sqrt 3.(\sqrt 3)^{n-1}=(\sqrt 3)^{n-1+1}=(\sqrt 3)^n$
$729=3^6$ and $\sqrt 3=3^{1/2}$
$\Rightarrow\:3^6=3^{n/2}$
Since the base on either side is equal (=3) the powe on either side is also equal.
$\Rightarrow\:6=\large\frac{n}{2}$ or $n=12$
$i.e., 12^{th}$ term of the given sequence is $729$
answered Feb 25, 2014 by rvidyagovindarajan_1
 

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