Question

Find the ratio of a6 and a8 in the geometric progression root 32,root 16,root 8.......​

Answer 1

Step-by-step explanation:

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Answer 2

The ratio of  $a_6$ and $a_8$ is equal to "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".