Question

insert five gm between 1/3 and 9 and verify that their products is the 5th power of the gm between 1/3 and 9

Answer 1

let 5 geometric means be a2,a3,a4,a5,a6 term of GP where a1=1/3 and a7 9 =aR^6

27=R^6

$R=\sqrt{3}$

THEREFORE  5 GM are $\sqrt{3} /3,1,\sqrt{3} ,3,3\sqrt{3}$[/tex]

Answer 2

Solution: Let five geometric mean between $\frac{1}{3} \text{and}  9$ be→ a, a r,a r², a r³ , a$r^4$.

So, $\frac{1}{3} ,a, a r, a r^{2} , a r^{3}, a r^4  ,a r^5= 9$ forms Geometric progression.

Common Ratio = Second Term ÷ First Term = $\frac{a}{\frac{1}{3}}=3a$

Also, $ar^5= 9\\a \times(3a)^5=9\\3^5\times a^6= 9\\ a = [\frac{3^2}{3^5}]^\frac{1}{6}=[\frac{1}{3^3}]^\frac{1}{6}=\frac{1}{\sqrt3}$

And, r= 3 a = 3 × $\frac{1}{\sqrt3}$ = √3

So, the five gm between $\frac{1}{3}$ and 9 is =

               $\frac{1}{\sqrt3},\frac{1}{\sqrt3}\times\sqrt{3} ,\frac{1}{\sqrt3}\times3, \frac{1}{\sqrt3}\times3\sqrt{3} ,\frac{1}{\sqrt3}\times 9 \text{which are } \frac{1}{\sqrt3},1,\sqrt{3}, 3,3\sqrt{3}$