Which term of the following sequences:
$\sqrt{3}, 3,3 \sqrt{3}, \ldots$ is $729 ?$
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 $729 .$
Let $a_1, a_2, a_3, \ldots$. be a $GP$ of increasing positive numbers. If the product of fourth and sixth terms is $9$ and the sum of fifth and seventh terms is $24 ,$ then $a_1 a_9+a_2 a_4 a_9+a_5+a_7$ is equal to $.........$.
How many terms of $G.P.$ $3,3^{2}, 3^{3}$... are needed to give the sum $120 ?$
Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+k, \in N$. If $c_2=5$ and $c_3=13 / 4$ then $\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)$ is equal to
If $a,\;b,\;c$ are in $A.P.$, then ${3^a},\;{3^b},\;{3^c}$ shall be in
The value of $0.\mathop {234}\limits^{\,\,\, \bullet \,\, \bullet } $ is