Insert two numbers between $3$ and $81$ so that the resulting sequence is $G.P.$
Let $G_{1}$ and $G_{2}$ be two numbers between $3$ and $81$ such that the series, $3, G_{1}, G_{2}, 81,$ forms a $G.P.$
Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
$\therefore 81=(3)(r)^{3}$
$\Rightarrow r^{3}=27$
$\therefore r=3$ (Talking real roots only)
For $r=3$
$G_{1}=a r=(3)(3)=9$
$G_{2}=a r^{2}=(3)(3)^{2}=27$
Thus, the required two numbers are $9$ and $27$
Find the sum to indicated number of terms in each of the geometric progressions in $\left.x^{3}, x^{5}, x^{7}, \ldots n \text { terms (if } x \neq\pm 1\right)$
If $(y - x),\,\,2(y - a)$ and $(y - z)$ are in $H.P.$, then $x - a,$ $y - a,$ $z - a$ are in
If $s$ is the sum of an infinite $G.P.$, the first term $a$ then the common ratio $r$ given by
Let for $n =1,2, \ldots \ldots, 50, S _{ a }$ be the sum of the infinite geometric progression whose first term is $n ^{2}$ and whose common ratio is $\frac{1}{(n+1)^{2}}$. Then the value of $\frac{1}{26}+\sum\limits_{n=1}^{50}\left(S_{n}+\frac{2}{n+1}-n-1\right)$ is equal to
The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive, then its $7^{th}$ term is