Question

36, ___, ___, 32/3 is a geometric sequence, with some terms missing. First, please solve for "r", the common ratio between each term. What is the value of r? Please enter your answer as a fraction, such as 6/7 or 1/5. *

Answer 1

`(2)/(3)`

Explanation:

Given:

36, ___, ___, 32/3 is a geometric sequence

We need to find common ratio, r.

Solution:

As here is geometric sequence:-

First term =
`ar` = 36

Fourth term =
`ar^(4)` =
`(32)/(3)`

`r=?`

We can write
`ar^(4)` as
`ar* r^(3)`

`( \ ar=36, \ given)`

`ar^(4) =(32)/(3) \ given`

`ar* r^(3)=(32)/(3)`

`36* r^(3) =(32)/(3)`

Dividing both sides by 36

`(36)/(36) * r^(3) =(32)/(3*36) \\\\ r^(3) =(32)/(108) \\Taking\ cube\ root\ both\ sides\\\sqrt[3]{r^(3) } =\sqrt[3]{(32)/(108) } \\ \\ r=\sqrt[3]{(8)/(27) } \\ \\ r=\sqrt[3]{(2*2*2)/(3*3*3) }\\ \\ r=(2)/(3)`

Thus, common ratio, r is
`(2)/(3)`

2nd term =
`a r^(2) =ar* r=36*(2)/(3) =24\\`

3rd term=
`ar^(3) =ar* r^(2) =36* ((2)/(3) )^(2) =36*=(4)/(9) =16`

Answer 2

The geometric sequence is

36,24,16,
`\frac{32}3`,.......

Explanation:

The first term of the g.s is 36.

The 4th term of the given sequence is
`(32)/(3)`.

The
`n^(th)` term of the geometric sequence is

`T_n=ar^(n-1)`

Where a is the first term of the geometric sequence and r is the geometric sequence.

Then 4th term of the sequence is

`T_4=ar^(4-1)`

`\Rightarrow \frac{32}3=36r^3`

`\Rightarrow r^3=(32)/(3* 36)`

`\Rightarrow r=\sqrt[3]{ (8)/(27)}`

`\Rightarrow r=(2)/(3)`

Then, second term of the sequence
`=36*( \frac 23)^(2-1)`

`=36* \frac 23`

=24

The third term of the sequence is
`=36*( \frac 23)^(3-1)`

`=36*( \frac 23)^2`

=16

The geometric sequence is

36,24,16,
`\frac{32}3`,.......