MCQ
The two geometric means between the numbers $1$ and $64$ are:
  • A
    $1$ and $64$
  • $4$ and $16$
  • C
    $2$ and $16$
  • D
    $8$ and $16.$

Answer

Correct option: B.
$4$ and $16$
Let the two $\text{G.M.s}$ between $1$ and $64$ be $G_1$ and $G_2$.
Thus, $1, \mathrm{G}_1, \mathrm{G}_2$ and $64$ are in $\text{G.P.}$
$64=1\times\text{r}^3$
$\Rightarrow\text{r}=\sqrt[3]{64}$
$\Rightarrow\text{r}=4$
$\Rightarrow\text{G}_1=\text{ar}=1\times4=4$
And, $\text{G}_2=\text{ar}^2=1\times4^2=16$
Thus, $4$ and $16$ are the required $\text{G.M.s.}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = } $
Let $x _{ i }(1 \leq i \leq 10)$ be ten observations of a random variable $X .$ If $\sum \limits_{ i =1}^{10}\left( x _{ i }- p \right)=3$ and $\sum \limits_{ i =1}^{10}\left( x _{ i }- p \right)^{2}=9$ where $0 \neq p \in R ,$ then the standard deviation of these observations is 
In the binomial $(2^{1/3} + 3^{-1/3})^n$, if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1/6$ , then $n =$
$cos^4 {\pi \over{8}}  + cos^4 {3\pi \over{8}}  +  cos^4 {5\pi \over{8}} +  cos^4 {7\pi \over{8}} = $ 
The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points $(a^2 + 1 , a^2 + 1 )$ and $(2a, - 2a)$, $a \ne 0$. Then for any $a$ , the orthocentre of this triangle lies on the line
If $5 + ix^3y^2$ and $x^3 + y^2 + 6i$ are conjugate complex numbers and arg $(x + iy) = \theta $ , then ${\tan ^2}\,\theta $ is equal to
If the vertices of triangle are $(0,2)$, $(1,0)$ and $(3,1)$, then the triangle is
$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $
Let ${\left( {1 + x} \right)^{10}} = \sum\limits_{r = 0}^{10} {{C_r}{x^r}} $ and ${\left( {1 + x} \right)^7} = \sum\limits_{r = 0}^7 {{d_r}{x^r}} $ . If $P = \sum\limits_{r = 0}^5 {{C_{2r}}} $ and $Q = \sum\limits_{r = 0}^3 {{d_{2r + 1}}} $ , then $\frac{P}{{2Q}}$ is equal to
A student is allowed to select at most $n$ books from a collection of $(2n + 1)$ books. If the total number of ways in which he can select one book is $63$, then the value of $n$ is