Question

Find the ratio of the a6 and a8 of Geometric progression root 32 and root 16 and root 8....

Answer 1

Answer:

geometric progression: root32, root16, root8.......

a1=root32

r=root16/root32

r=1/root2

a6=a+r^[n-1]

a6=root32*[1/root2]^5

a6=1

a8=root32*[1/root2]^7

a8=1/2

Step-by-step explanation:

Answer 2

The ratio of  $a_6$ and $a_8$ = 2 : 1

Step-by-step explanation:

The given geometric progression (GP) are:

$\sqrt{32} ,\sqrt{16} ,\sqrt{8} ,......$

$= 4\sqrt{2}, 4, 2\sqrt{2}, ........$

Here, first term (a) = $4\sqrt{2}$ and common ratio (r) = $\dfrac{4}{4\sqrt{2}} =\dfrac{1}{\sqrt{2}}$

To find, the ratio of  $a_6$ and $a_8$ = ?

We know that,

The nth term of geometric progression(GP)

$a_{n} =ar^{n-1}$

∴ The 6th term of geometric progression(GP)

$a_{6} =(4\sqrt{2} )(\dfrac{1}{\sqrt{2} } )^{6-1}$

$=(4\sqrt{2} )(\dfrac{1}{\sqrt{2} } )^{5}=(4\sqrt{2} )\dfrac{1}{4\sqrt{2} }=1$

The 8th term of geometric progression(GP)

$a_{8} =(4\sqrt{2} )(\dfrac{1}{\sqrt{2} } )^{8-1}$

$=(4\sqrt{2} )(\dfrac{1}{\sqrt{2} } )^{7}=(4\sqrt{2} )\dfrac{1}{8\sqrt{2} }=\dfrac{1}{2}$

∴ The ratio of  $a_6$ and $a_8$ = 1 : $\dfrac{1}{2}$ = 2 : 1

Thus, the ratio of  $a_6$ and $a_8$ is equal to "2 : 1".