Question

Insert four G.M's between 2 and 486.

Answer 1

Hint: First let us suppose G.M's be $2,{a_1},{a_2},{a_3},{a_4},486$. Hence these term are in GP we know the sixth term of this GP that is equal to $486$, mean that $a{r^5} = 486$ where $a$ = first term of GP equal to $2$ from here we get the value of $r$ then we got the remaining term as ${a_1} = ar,{a_2} = a{r^2},{a_3} = a{r^3},{a_4} = a{r^4}$

Complete step by step answer:
As in the question we have to insert four geometric means between this $2$ and $486$.
Geometric mean means we have to add four numbers in between $2$ and $486$ so that the series form will be in G.P.
lets us take G.M's be $2,{a_1},{a_2},{a_3},{a_4},486$ so these term is in G.P.
If we write the general \[6\] term of GP then it can be written as $a,ar,a{r^2},a{r^3},a{r^4},a{r^5}$ where a = first term of GP while r = common ratio of the GP.
For the given question we know $a = 2$ and we know that the last term $a{r^5}$ is equal to $486$ Now we have to find the common ratio of GP for which we ,
$\Rightarrow a{r^5} = 486$
or by putting the value of a , we get
$\Rightarrow 2{r^5} = 486$
$\Rightarrow {r^5} = 243$
or $243$ is the ${5^{th}}$ power of $3$ hence ,
$\Rightarrow {r^5} = {3^5}$
on comparing both side we get $r = 3$
Hence ${a_1} = ar,{a_2} = a{r^2},{a_3} = a{r^3},{a_4} = a{r^4}$
by putting the value of $a = 2$ and $r = 3$ we get ,
$\Rightarrow {a_1} = 2 \times 3,{a_2} = 2 \times {3^2},{a_3} = 2 \times {3^3},{a_4} = 2 \times {3^4}$
by solving the further we get
$\Rightarrow {a_1} = 6,{a_2} = 18,{a_3} = 54,{a_4} = 162$

$\therefore $ Hence the four geometric mean will be $6,18,54,162$ in between $2$ and $486$.

Note:
In between the question we will compare $a{r^5} = 486$ for finding the common ratio of the GP we will also use the simple formula for the common ratio that is $r = \left( {\dfrac{b}{a}} \right)\dfrac{1}{{n + 1}}$ where $a$= first term $b$ = last term of GP and $n$ = number of term which we have to insert in between the GP.
Similarly, if we have to find AM between two number we use $d = \dfrac{{b - a}}{{n + 1}}$ where a= first term b = last term of AP and n is the number of terms we have to insert and then we get d from which we get next numbers in between.