Question

Insert three numbers between $1$ and $256$ so that the resulting sequence is a $\text{G.P}$.

Answer 1

Let $G_1,G_2,G_3$ be three numbers such that $1,G_1,G_2,G_3,256$ is $\text{G.P}$
Here, $a=1,b=256$ and $n=5$.

Step $1:$ Finding the value of common ratio $r$.
We know the $n^{th}$ term of $\text{G.P}$ is given by
$b=a{r}^{n-1}$.
$\Rightarrow 256=1\cdot r^{5-1}$
$\Rightarrow r^4=256$
$\Rightarrow r^4=(\pm 4{)}^4$
$\therefore r=4$ or $-4$

Step $2:$ Finding inserted three numbers.
Case $1:$ when $a=1$ and $r=4$, then inserted numbers are,

$G_1=ar=1\times 4=4$
$G_2=a{r}^2=1\times 4^2=16$
$G_3=a{r}^3=1\times 4^3=64$

Case $2:$ when $a=1$ and $r=-4$, then inserted numbers are,

$G_1=ar=1\times (-4)=-4$
$G_2=a{r}^2=1\times (-4{)}^2=16$
$G_3=a{r}^3=1\times (-4{)}^3=-64$

Hence, the inserted numbers between $1$ and $256$ are $4,16,64$ for $r=4$ and $-4,16,-64$ for $r=-4$.