- Home
- Standard 11
- Mathematics
8. Sequences and Series
easy
શ્રેણી $\sqrt{3}, 3,3 \sqrt{3}, \ldots$ નું કેટલામું પદ $729$ થાય ?
A
$12^{\text {th }}$
B
$12^{\text {th }}$
C
$12^{\text {th }}$
D
$12^{\text {th }}$
Solution
The given sequence is $\sqrt{3}, 3,3 \sqrt{3,}, \ldots \ldots$
$a=\sqrt{3}$ and $r=\frac{3}{\sqrt{3}}=\sqrt{3}$
Let the $n^{\text {th }}$ term of the given sequence be $729 .$
$a_{n}=a r^{n-1}$
$\therefore a r^{n-1}=729$
$\Rightarrow(\sqrt{3})(\sqrt{3})^{n-1}=729$
$\Rightarrow(3)^{1 / 2}(3)^{\frac{n-1}{2}}=(3)^{6}$
$\Rightarrow(3)^{\frac{1}{2}+\frac{n-1}{2}}=(3)^{6}$
$\therefore \frac{1}{2}+\frac{n-1}{2}=6$
$\Rightarrow \frac{1+n-1}{2}=6$
$\Rightarrow n=12$
Thus, the $12^{\text {th }}$ term of the given sequence is $729 .$
Standard 11
Mathematics
