Question

Four geometric means are inserted between 1/8 and 128. Find the third geometric mean.

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

The third geometric mean=8

Step-by-step explanation:

Let four geometric  $a_1,a_2,a_3,$ and $a_4$ are inserted between 1/8 and 128.

It forms a G.P

Let r be the common ratio between two consecutive terms of given sequence.

$a_1=a=\frac{1}{8}$

$a_6=128$

nth term of G.P

$a_n=ar^{n-1}$

$a_6=ar^5$

Substitute the values then we get

$128=\frac{1}{8}r^5$

$r^5=128\times 8=1024$

$r=\sqrt[5]{1024}=4$

$a_4=ar^3$

Substitute the values then we get

$a_4=\frac{1}{8}(4)^3=8$

Hence, the third geometric mean=8

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